Question XIII. Consider each numbered statement (133-140) concerning.pdf
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Question XIII. Consider each numbered statement (133-140) concerning various immunological facts, concepts and terminology as a completely independent assertion. Using your knowledge, please, assess these statements as factually correct (= true, T) or incorrect (= false, F): 133. The 12/23 rule means that a diversity (D) segment is almost always included in the formation of the immunoglobulin heavy-chain V region. 134. Although antigen-antibody reactions occur via non-covalent bonds, they are by their nature irreversible. 135. The T-cell receptor-encoding genomic sequences in our circulating T cells are the same as those our parents carry in their T cells. 136. Alternative splicing of RNA is utilized by T cells to form the and chains of a T-cell receptor. 137. In generating a B-cell receptor gene, Vk segments sometimes join to Cl segments. 138. In generating a T-cell receptor gene, Va segments sometimes join to Cd segments. 139. The switch in constant region usage from IgM to another immunoglobulin class (= isotype switching) is always mediated by DNA rearrangements. 140. A major reason why one individual might mount a response against a specific self-peptide, while other individuals might not is that there are differences in peptide binding potential of different MHC molecules. Solution 133. The 12/23 rule means that a diversity (D) segment is almost always included in the formation of the immunoglobulin heavy-chain V region. False. Recombination only occurs between gene segments located on the same chromosome. It generally follows the rule that only a gene segment flanked by a RSS with a 12-base pair (bp) spacer can be joined to one flanked by a 23 bp spacer RSS. This is known as the 12/23 rule. Thus, for the heavy chain, a DH gene segment can be joined to a JH gene segment and a VH gene segment to a DH gene segment, but VH gene segments cannot be joined to JH gene segments directly, as both VH and JH gene segments are flanked by 23 bp spacers and the DH gene segments have 12 bp spacers on both sides . So it can be included but not always. 134. Although antigen-antibody reactions occur via non-covalent bonds, they are by their nature irreversible. False. They are reversible. 135. The T-cell receptor-encoding genomic sequences in our circulating T cells are the same as those our parents carry in their T cells. False, The genes for encoding T-cell receptor are rearranged and edited during T-cell development in the thymus so that each mature T cell carries a different T-cell receptor gene sequence. 136. Alternative splicing of RNA is utilized by T cells to form the and chains of a T-cell receptor. False.
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1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic information? How are these similar
and how are they different? How can they be used to map bacterial genomes?
b) Explain how genes in phages are mapped. (This is different from the above question)
Solution
Ques-1: What are three ways that bacteria can exchange genetic information? How are these
similar and how are they different? How can they be used to map bacterial genomes?
Answer:
Horizontal gene transfer (HGT) or lateral gene transfer is the transfer of genes between two
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their evolution. HGT accelerates evolution.
Horizontal gene transfer: 1. conjugation, 2. transformation; 3. transduction
Similarities & differences:
Horizontal gene transfer is the transfer of genomes by between bacteria by the routes that does
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three mechanism of horizontal gene transfer.
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transduction and abolish transfer of chromosomal DNA from one bacterial cell to another by
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bacterial membrane to lyse finally incorporate phage genome into the bacterial cell for
integration. Therefore, for efficient transduction, presence of recipient cell- surface receptors is
essential in the culture otherwise, bacteria exert resistant effects to the phages.
Transformation type of gene transfer occurs between bacteria. The genes released by dead
bacteria can be transferred to the live bacteria, and get integrated into its genome, which results
in genetic recombination of the host.
Transduction is also involving the genetic material transfer, but this needs bacteriophage as
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capsids of some of these virions carry the bacterial DNA fragments into the new host bacteria,
followed by genetic recombination of the host bacteria.
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bridge like structure. Bacterial conjugation needs the conjugation tube for genome transfer, both
transduction and transformation does not need this.
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On a Sunday in April, dog bite victims arrive at Carver Memorial Hospital at a historical rate of
0.6 victim per day. On a given Sunday in April, what is the probability that exactly two dog bite
victims will arrive?
Solution
Given X~Poisson(=0.6)
P(X=x)=(0.6^x)*exp(-0.6)/x!
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Molecules like glucose usually need a special transporter protein to provide a tunnel through cell
membranes. Explain why using the chemical properties of the cell membrane.
Solution
Hi,
The cell envelope is made up of lipid bilayer. This bilayer is hydrophobic in nature ( non polar)
and does not allow all molecules to pass freely across the membrane. Only small uncharged
molecules ( O2, CO2) can pass through it. Even the small uncharged polar molecules such as
H2O can pass through it it. But the larger uncharged polar molecules cannot pass through it
freely, such as Glucose.
The charged molecules such as Na+, H+ are polar in nature and hence cannot pass through the
non-polar lipid bilayer regardless of size. They all need specialized transporters to facilitate their
transport without interaction with hydrophobic bilayer..
More Related Content
Similar to Question XIII. Consider each numbered statement (133-140) concerning.pdf
Organization and expression of Ig genes
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This document discusses the structure, properties, genetic organization, and genetic rearrangement of antibodies which allows for antibody diversity. It notes that antibodies are glycoproteins produced by plasma cells that recognize and bind antigens. Their genetic organization involves genes for heavy and light chains being located on different chromosomes, with multiple gene segments that undergo rearrangement. This rearrangement at the DNA and RNA level involves joining of V, D, and J genes for heavy chains and V and J genes for light chains, and splicing with C region genes to produce different antibody classes. This genetic rearrangement and splicing is essential for producing the antibody diversity needed to recognize a wide range of antigens.
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This document discusses the genetics of antibody diversity. It describes three main mechanisms that generate antibody diversity: somatic recombination, somatic hypermutation, and class switching. Somatic recombination involves randomly combining variable, diversity, and joining gene segments to generate a wide variety of antibody sequences. Somatic hypermutation further introduces random mutations during B cell activation, allowing higher affinity antibodies to be selected. Class switching changes the constant region of the antibody to alter effector functions. These three processes together allow the immune system to generate a vast repertoire of antibodies with unique specificities.
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This presentation discusses the concept of the generation of antibody diversity through immunoglobulin(Ig) gene rearrangements. The mechanism of V-J rearrangements (light chain) and V-D-J rearrangements (heavy chain) is elaborated in detail along with the Ig constant region rearrangements, called class-switching. The factors dictating antibody diversity and antigen-antibody interactions have also been discussed briefly.
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REGULATION OF Ig GENE TRANSCRIPTION
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The document summarizes key aspects of T cell antigen receptor (TcR) structure and generation of diversity. It describes how TcR were discovered using monoclonal antibodies that recognize unique structures on T cell clones. TcR are heterodimers composed of α and β chains that are similar to antibody structures but do not undergo somatic hypermutation. TcR diversity is generated through combinatorial rearrangement of variable (V), diversity (D), and joining (J) gene segments, as well as junctional diversity from imprecise joining and addition of untemplated nucleotides.
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T cells recognize antigens presented by MHC molecules on other cells. Antigens from intracellular sources such as viruses are processed and presented by MHC class I to CD8+ T cells. Antigens from extracellular sources such as bacteria that are endocytosed are processed and presented by MHC class II to CD4+ T cells. The antigen processing pathways and the classes of MHC molecules that present the peptides allow the immune system to mount appropriate responses targeting different types of pathogens.
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Antibody
- Antibodies are Y-shaped proteins produced by B cells that recognize and bind to pathogens. They have great diversity and specificity that allows them to target a wide range of antigens.
- Antibodies are composed of two heavy chains and two light chains that form an antigen binding site. The variable regions of the heavy and light chains are responsible for binding to different antigens.
- Monoclonal antibodies are produced from a single clone of B cells and have identical antigen specificity, allowing them to be used as standardized reagents for research and medical purposes. They are produced by fusing B cells with myeloma cells to form immortal hybridomas.
1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic information? How are these similar
and how are they different? How can they be used to map bacterial genomes?
b) Explain how genes in phages are mapped. (This is different from the above question)
Solution
Ques-1: What are three ways that bacteria can exchange genetic information? How are these
similar and how are they different? How can they be used to map bacterial genomes?
Answer:
Horizontal gene transfer (HGT) or lateral gene transfer is the transfer of genes between two
organisms that are phylogenetically unrelated but can readily exchange genes by conjugation or
transduction or transformation. HGT commonly occurs in prokaryotes and it plays a key role in
their evolution. HGT accelerates evolution.
Horizontal gene transfer: 1. conjugation, 2. transformation; 3. transduction
Similarities & differences:
Horizontal gene transfer is the transfer of genomes by between bacteria by the routes that does
not involve parent-offspring transmission. Conjugation, transduction and transformation are the
three mechanism of horizontal gene transfer.
In the above case, it has clearly illustrated that the conditions, which can successfully prevents
transduction and abolish transfer of chromosomal DNA from one bacterial cell to another by
generalized transduction with P22 bacteriophage. Initially recipient cell- surface receptors are
important for the attachment of the bacteriophage (P22) and finally spikes of phage enable
bacterial membrane to lyse finally incorporate phage genome into the bacterial cell for
integration. Therefore, for efficient transduction, presence of recipient cell- surface receptors is
essential in the culture otherwise, bacteria exert resistant effects to the phages.
Transformation type of gene transfer occurs between bacteria. The genes released by dead
bacteria can be transferred to the live bacteria, and get integrated into its genome, which results
in genetic recombination of the host.
Transduction is also involving the genetic material transfer, but this needs bacteriophage as
media. The bacteriophage infects the bacteria in lytic cycle and releases the new virions, the
capsids of some of these virions carry the bacterial DNA fragments into the new host bacteria,
followed by genetic recombination of the host bacteria.
Conjugation: It is the transfer of genetic material from cell to cell by means of direct contact or a
bridge like structure. Bacterial conjugation needs the conjugation tube for genome transfer, both
transduction and transformation does not need this.
Mapping the genome -mechanisms
F factor is mainly pertaining to a class of conjugate plasmids, which meticulously originated to
control sexual functions in prokaryotes via fertility inhibition. The origin of F-factor is mainly
with most common functional segments such as gene traJ, Ori, OriC. The structure of this F
factor contains OriT for origin of transfer and acts as a starting point for gene transmission.
T Cell Antigen Receptor
The document summarizes key aspects of T cell antigen receptor (TcR) discovery and structure:
1) TcR were discovered through making monoclonal antibodies against the antigen receptor of cloned T cells, which led to identifying the TcR alpha and beta chains.
2) TcR genes rearrange through somatic recombination similar to immunoglobulin genes, generating diversity through combinations of variable, diversity and joining gene segments.
3) Junctional diversity is introduced through addition of non-templated and palindromic nucleotides during the recombination process.
Recombination Final.ppt
Recombination
Breaking and rejoining of two parental DNA molecules to produce new DNA molecules
Types of recombination
Definition of recombination
Gene Conversion – Characteristics
Holliday model
Holliday junction cleavage
HLA Matching
This document discusses stem cell transplantation and histocompatibility. It explains that stem cell transplants can be rejected by the recipient's immune system, so the transplanted cells must closely match the recipient to avoid being recognized as foreign. To determine if a donor is a good immunological match, a tissue typing test identifies HLA antigens on immune cells; identical HLA antigens indicate a good match. The document provides background on HLA antigens and their role in the immune response and histocompatibility in stem cell transplantation.
Complementarity Determining Regions
This presentation aims to describe the variability present in antibodies.what the Ig superfamily have in common and the various functions it performs.Role of different enzymes imparting diversity to the variable region has been covered.
Lecture 7 - 10.12.09 - DNA Recombination.ppt
This document discusses mechanisms of genetic recombination and antigenic variation. It begins by explaining the benefits of recombination, including separating favorable and unfavorable mutations and generating new gene combinations. It then describes several types of homologous recombination, including double-strand break repair and gene conversion. Specific proteins involved in initiating recombination, such as RecBCD and RecA, are also discussed. The document focuses on examples of antigenic variation in pathogenic protozoa, including Plasmodium falciparum and African trypanosomes. It describes how these organisms use large gene families and recombination-mediated switching of variant surface antigens to evade the immune system through clonal antigenic variation.
HLA
The document discusses HLA typing and its implications. It begins with a brief history of the discovery of the major histocompatibility complex (MHC) in mice and humans. It then describes the structure and functions of the MHC, including the class I, class II, and class III regions. The document discusses methods of HLA typing, including serological testing using microcytotoxicity and molecular methods. It notes some implications of HLA typing, such as its role in organ transplant matching and susceptibility to autoimmune disease.
bb_unit7RhSpring2011.ppt
The document discusses the Rh blood group system, which is one of the most complex blood group systems. It contains over 50 antigens, but focuses on the five main antigens: D, C, c, E, and e. The Rh system is clinically significant as Rh antibodies can cause hemolytic disease of the newborn. The document compares three systems of Rh nomenclature: Fisher-Race, Wiener, and Rosenfield/ISBT. It also discusses Rh phenotyping and genotyping, weak D expression, and inheritance patterns of Rh antigens.
T CELL RECEPTOR.pptx
The T cell receptor (TCR) is a protein complex found on the surface of T cells that is responsible for recognizing fragments of antigen bound to MHC molecules. It consists of an alpha and beta chain, with 95% of T cells containing these chains and 5% containing gamma and delta chains instead. Each chain contains a variable region that binds the peptide-MHC complex and a constant region near the cell membrane. The variable regions contain three hypervariable complementarity-determining regions important for antigen recognition. The TCR is associated with CD3 proteins that transmit activation signals into the T cell upon peptide binding. TCR diversity arises from genetic recombination of DNA segments during T cell development.
Similar to Question XIII. Consider each numbered statement (133-140) concerning.pdf (20)
V(D)J rearrangements and Antigen Antibody interactions
V(D)J rearrangements and Antigen Antibody interactions
Generation of Antibody Diversity- Quick revision from Kuby through presentation
Generation of Antibody Diversity- Quick revision from Kuby through presentation
1a- What are three ways that bacteria can exchange genetic informati.pdf
1a- What are three ways that bacteria can exchange genetic informati.pdf
More from fasttrackscardecors
On a Sunday in April, dog bite victims arrive at Carver Memorial Hos.pdf
On a Sunday in April, dog bite victims arrive at Carver Memorial Hospital at a historical rate of
0.6 victim per day. On a given Sunday in April, what is the probability that exactly two dog bite
victims will arrive?
Solution
Given X~Poisson(=0.6)
P(X=x)=(0.6^x)*exp(-0.6)/x!
So the probability is
P(X=2)=(0.6^2)*exp(-0.6)/2=0.09878609.
Molecules like glucose usually need a special transporter protein to.pdf
Molecules like glucose usually need a special transporter protein to provide a tunnel through cell
membranes. Explain why using the chemical properties of the cell membrane.
Solution
Hi,
The cell envelope is made up of lipid bilayer. This bilayer is hydrophobic in nature ( non polar)
and does not allow all molecules to pass freely across the membrane. Only small uncharged
molecules ( O2, CO2) can pass through it. Even the small uncharged polar molecules such as
H2O can pass through it it. But the larger uncharged polar molecules cannot pass through it
freely, such as Glucose.
The charged molecules such as Na+, H+ are polar in nature and hence cannot pass through the
non-polar lipid bilayer regardless of size. They all need specialized transporters to facilitate their
transport without interaction with hydrophobic bilayer..
Missy Crane opened a public relations firm called Goth on August 1, 2.pdf
Missy Crane opened a public relations firm called Goth on August 1, 2018 The During
Septembe following amounts summarize her business on August 31, 2018 2018, the business
completed the following transactions 1 (Click the icon to view the transactions.) EE (click the
icon to view the amounts) Analyze the effects of the transactions on the accounting equation of
Goth. Analyze the events chronologically, one transaction at a time. Beginning with transaction
a, calculate the balance in each account after analyzing the effect of the transaction on the
accounting equation. (Complete onily the necessary answer boxes for your transaction lines [Do
not enter any zeros for your transaction lines ] Carry down all balances to the \"Bal.\" line,
including zero balance accounts, entening a \"O\" for any zero balances Enter a decrease in an
account with a minus sign or parentheses. Abbreviations used AP Accounts Payable /R -
Accounts Receivable: Adv Advertising, Cap -Crane, Capital; Exp Expense, Liab Liabilities Rev.
Revenue, Sup = Supplies withdr = Crane, withdrawals ) ASSETS LIAB.+ EQUITY Cash + A/R
+ Office + Land = A/P + Cap Service Rent Adv Sup RevEXp 3,400 Exp Bal 2300 3.400 12.000
= 8,0006 300 0 Bal
Solution
Assets
=
Liabilities
+
Equity
cash
+
A/r
+
Office supplies
+
land
=
A/p
+
cap
-
withdrawls
+
Service revenue
-
Rent expense
-
Adve expense
bal
2300
3400
0
12000
8000
6300
0
3400
0
0
a
16000
16000
bal
18300
3400
0
12000
8000
22300
0
3400
0
0
b
1600
1600
bal
19900
3400
0
12000
8000
22300
0
5000
0
0
c
(8000)
(8000)
bal
11900
3400
0
12000
0
22300
0
5000
0
0
d
600
600
bal
11900
3400
600
12000
600
22300
0
5000
0
0
e
2100
(2100)
bal
14000
1300
0
12000
0
22300
0
5000
0
0
f
(1100)
1100
bal
12900
1300
0
12000
0
22300
1100
5000
0
0
g
5000
5000
bal
12900
6300
0
12000
0
22300
1100
10000
0
0
h.1
(1100)
1100
h.2
(450)
450
bal
11350
6300
0
12000
0
22300
1100
10000
1100
450
Journal entries
Capital Cr. 16000
Service revenue Cr. 1600
Cash Cr. 8000
Accounts payable Cr. 600
A/R Cr. 2100
Cash Cr. 1100
Service revenue Cr. 5000
Cash Cr. 1100
2. Advertising expense Dr. 450
Cash Cr. 450
Assets
=
Liabilities
+
Equity
cash
+
A/r
+
Office supplies
+
land
=
A/p
+
cap
-
withdrawls
+
Service revenue
-
Rent expense
-
Adve expense
bal
2300
3400
0
12000
8000
6300
0
3400
0
0
a
16000
16000
bal
18300
3400
0
12000
8000
22300
0
3400
0
0
b
1600
1600
bal
19900
3400
0
12000
8000
22300
0
5000
0
0
c
(8000)
(8000)
bal
11900
3400
0
12000
0
22300
0
5000
0
0
d
600
600
bal
11900
3400
600
12000
600
22300
0
5000
0
0
e
2100
(2100)
bal
14000
1300
0
12000
0
22300
0
5000
0
0
f
(1100)
1100
bal
12900
1300
0
12000
0
22300
1100
5000
0
0
g
5000
5000
bal
12900
6300
0
12000
0
22300
1100
10000
0
0
h.1
(1100)
1100
h.2
(450)
450
bal
11350
6300
0
12000
0
22300
1100
10000
1100
450.
Justify that the following circuit represents the functionality of a .pdf
Justify that the following circuit represents the functionality of a Full-Adder circuit.
Solution
1) Sender, receiver, medium, message, and protocol.
2) Internet draft A working Internet document (a work in progress) with no official status and a
six-month lifetime.
Proposed Standard: It was a standard developed by IETF as a less mature and well defined
standard.
3) a. If one connection fails, the other connections will still be working.
b. The other devices will still be able to send data through the hub; there will be no
access to the device which has the failed connection to the hub.
c. All transmission stops.
d. The failed connection may disable the whole network unless it is a dual ring.
4) Application Layer: Packet
Network Layer: IP Datagram
DataLink Layer: Frame
5) Application Layer: sagarsetti4u@gmail.com
Network Layer:Logical Addresses
Data Link Layer: Link Layer Addresses
6) No, multiplexing/ de-multiplexing at transport layer does notmean combining several upper-
layer packets (from the sameor different applications) into one transport-layer packet. Itonly
means that each of the transport-layer protocols such asTCP can carry a packet from any
application-layer protocolthat needs its service. However, a transport-layer packet cancarry one,
and only one, packet from an application layerprotocol. For instance, UDP can carry a message
from FTPin one user datagram and a message from HTTP in anotheruser datagram
7) a) Print your Boarding Pass and check in your Luggage.
b) Head to Security
c) Find your gate/terminal
d) Wait for Your Plane
e) Wait for Your Announcement
f) Get the Boarding pass checked
g) Enter the aircraft
h) Get Settled Down
We can Prove that it is a full adder by constructing Truth Table.
Initially Full adder Truth Table is given by
x
y
z
Sum
Carry
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
1
0
0
1
0
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1
Now Let us See For (xy) Cin and For (xy) Cin+xy
x
y
Cin
(xy)
(xy) Cin
(xy) Cin
xy
(xy) Cin+xy
0
0
0
0
0
0
0
0
0
0
1
0
1
0
0
0
0
1
0
1
1
0
0
0
0
1
1
1
0
1
0
1
1
0
0
1
1
0
0
0
1
0
1
1
0
1
0
1
1
1
0
0
0
0
1
1
1
1
1
0
1
0
1
1
Hence we can see that (xy) Cin and Sum are identical.
(xy) Cin+xy and Carry are Identical.
Hence it a Full Adder
x
y
z
Sum
Carry
0
0
0
0
0
0
0
1
1
0
0
1
0
1
0
0
1
1
0
1
1
0
0
1
0
1
0
1
0
1
1
1
0
0
1
1
1
1
1
1.
Inventor classwhat is the purpose of using assemply constraints.pdf
Inventor class
what is the purpose of using assemply constraints?
Solution
You use assembly constraints and the purpose is to create parametric relationships between parts
in the assembly. Just as you use 2D constraints to control 2D geometry, you use 3D constraints
in an assembly to position parts in
relation to other parts.
IncompressibilityI cant understand some parts in a book.The .pdf
Incompressibility
I can\'t understand some parts in a book.
\"The pointwise incompressibility condition i.s. the condition of vanishing dilatational strain,
provides certain constraints to the assumed displacement finite element formulation. As an
example the displacements for a four node rectangular plane strain element can be assumed as
These three conditions of constraint will limit the element motion to only two possible
deformation modes which, of course, cannot represent more general state of strain.\"
My questions: I understand that these constraints limit the element motion, but why just two
possible deformation modes are available?
How can I know which of them?
Solution.
If you were to illuminate an Elodea leaf with light at 550 nm, would.pdf
If you were to illuminate an Elodea leaf with light at 550 nm, would you expect CO2
consumption by the leaf? Why?
Solution
Answer:
CO2 consumption by the Elodea leaf depends on the rate of photosynthesis. The light at 550nm
corresponds to green light which is not favourable for photosynthesis. The chlorophyll pigment
reflects green light and absorbs other wavelengths of light between 400-700nm for
photosynthesis. Se we expect less consumption of CO2 at 550nm.
Identify the three major types of controls that organizations can us.pdf
Identify the three major types of controls that organizations can use to protect their information
resources ?
Solution
1. At secure organizations, information security is supported by senior management. Support
includes making resources and budget available for information security, as well as clear
statements by senior management that information security is a priority for the organization.
Since senior managers establish priorities and set the tone for an organization, it is difficult to be
a secure organization without their clear and consistent support. As a result of the recent spate of
high-profile security breaches, most senior managers now understand the importance of
information security and will support information security efforts.
2. Secure organizations regularly identify and document how sensitive data --customer and/or
proprietary -- flows in, through and out of the organization. This enables an organization to focus
its time, effort and money on protecting its sensitive data. Conversely, it\'s difficult for an
organization to protect what it doesn\'t know about, and organizations struggle to protect their
data if they don\'t perform this exercise
3. Secure organizations create and maintain a formal, documented inventory of all systems that
process, transmit or store sensitive data -- including the operating system, if it\'s physical or
virtualized, and what major applications have been installed. Without such an inventory, an
organization can\'t fully understand what systems it must protect. Having such an inventory
allows an organization to quickly determine whether a particular security vulnerability is relevant
to the organization\'s systems.
4. Secure organizations segment sensitive systems from non-sensitive systems through jump
servers, firewall rules, router ACLs or switch VLANs. This minimizes the attack surface for an
organization\'s sensitive systems and allows access to the systems to be tightly controlled and
logged.
5. Secure organizations have a strong change-control process that is rigorously enforced.
Changes, including emergency changes, are fully documented then formally reviewed and
approved. Unapproved changes can lead to security vulnerabilities that nobody knows about
until there\'s a breach.
6. Secure organizations have a strong configuration management process. Sensitive systems are
hardened and built only with necessary functionality via an automated build process or a
managed configuration software tool such as Puppet or Chef. After the initial build,
configuration software tools, which regularly check the configuration of systems, are used to
ensure systems stay hardened or strong change control is used to maintain system configuration
and prevent server creep..
I have C++ question that I do not know how to do, Can you teach me t.pdf
I have C++ question that I do not know how to do, Can you teach me this question:
Draw a CLASS diagram for Binary Search Tree (BTS) :
can you explain me each step, thank you
Solution
//program to convert Binary Tree to Binary Search Tree
#include
#include
// A binary tree node structure
struct node
{
int data;
struct node *left;
struct node *right;
};
void storeInorder (struct node* node, int inorder[], int *index_ptr)
{
// Base Case
if (node == NULL)
return;
// first store the left subtree
storeInorder (node->left, inorder, index_ptr);
// Copy the root\'s data
inorder[*index_ptr] = node->data;
(*index_ptr)++; // increase index for next entry
// finally store the right subtree
storeInorder (node->right, inorder, index_ptr);
}
/* A helper function to count nodes in a Binary Tree */
int countNodes (struct node* root)
{
if (root == NULL)
return 0;
return countNodes (root->left) +
countNodes (root->right) + 1;
}
// Following function is needed for library function qsort()
int compare (const void * a, const void * b)
{
return ( *(int*)a - *(int*)b );
}
/* A helper function that copies contents of arr[] to Binary Tree.
This functon basically does Inorder traversal of Binary Tree and
one by one copy arr[] elements to Binary Tree nodes */
void arrayToBST (int *arr, struct node* root, int *index_ptr)
{
// Base Case
if (root == NULL)
return;
/* first update the left subtree */
arrayToBST (arr, root->left, index_ptr);
/* Now update root\'s data and increment index */
root->data = arr[*index_ptr];
(*index_ptr)++;
/* finally update the right subtree */
arrayToBST (arr, root->right, index_ptr);
}
// This function converts a given Binary Tree to BST
void binaryTreeToBST (struct node *root)
{
// base case: tree is empty
if(root == NULL)
return;
/* Count the number of nodes in Binary Tree so that
we know the size of temporary array to be created */
int n = countNodes (root);
// Create a temp array arr[] and store inorder traversal of tree in arr[]
int *arr = new int[n];
int i = 0;
storeInorder (root, arr, &i);
// Sort the array using library function for quick sort
qsort (arr, n, sizeof(arr[0]), compare);
// Copy array elements back to Binary Tree
i = 0;
arrayToBST (arr, root, &i);
// delete dynamically allocated memory to avoid meory leak
delete [] arr;
}
/* Utility function to create a new Binary Tree node */
struct node* newNode (int data)
{
struct node *temp = new struct node;
temp->data = data;
temp->left = NULL;
temp->right = NULL;
return temp;
}
/* Utility function to print inorder traversal of Binary Tree */
void printInorder (struct node* node)
{
if (node == NULL)
return;
/* first recur on left child */
printInorder (node->left);
/* then print the data of node */
printf(\"%d \", node->data);
/* now recur on right child */
printInorder (node->right);
}
/* Driver function to test above functions */
int main()
{
struct node *root = NULL;
/* Constructing tree given in the above figure
10
/ \\
30 15
/ \\
20 5 */
root = newNode(10);
root->left = newNode(30.
I need proper and details explanation for this case study Financial .pdf
I need proper and details explanation for this case study Financial statement part (i, ii & iii Area
of concern)
Analyse the financial statements that have been prepared by Home Range Ltd’s financial
director. In particular, comment on the following aspects of the company:
i. Areas of concern in financial performance, focusing mainly on information from the income
statement
ii. Areas of concern in financial health, focusing on ratios dependent on the income statement
and the balance sheet
iii. Areas of concern in cash flow management, focusing on information available from the cash
flow statement.
Assess the value of Clare Lombardo’s parent’s recent concern that ‘the company has not spent its
significant cash resources wisely in the past year’ (the year to 31 December 2017). What
practical steps should Clare Lombardo now undertake in order to help to maximise the value of
the business for resale in about 3 years?
Case study
Home Range Ltd
Clare Lombardo comes from a family that has bought, renovated and sold or rented houses as a
side-line to their main forms of employment for a number of years. This family interest has
proved to be a profitable small business. Her father is an electrician by trade but has developed
skills in other areas such as basic plumbing and carpentry. This has greatly reduced the cost of
‘doing up’ old properties. Clare’s mother works as a part-time bookkeeper so she has enough
spare time to keep proper financial records for the business.
Clare’s favourite subject at school was Art. While still at school Clare discovered a knack for
helping her parents to tastefully furnish renovated properties in preparation for their sale or rent.
After school Clare studied interior design while helping run her parent’s business.
About 20 years ago Clare was left money by a grandparent. She used this to open her first Home
Range shop in an exclusive part of London. It sells high quality home furnishings and
accessories to mainly wealthy customers who have limited time.
Clare’s luck, good taste and business acumen has proved very successful. She formed the private
company Home Range Ltd 18 years ago, and it has grown to consist of a chain of 25 stores in
wealthy parts of London and the south east of England. The retail outlets have been
complimented by an effective and profitable website.
Clare is the majority shareholder while her parents own the remaining shares in the business.
Clare has recently decided she wants to sell the business in about three years. She is conscious
that her parents are growing old and she also want to spend more time with her young family.
Clare’s parents have decided to retire from formal employment in the next few months. They
will still run their property business. While they are essentially passive shareholders in Home
Range Ltd, Clare still relies on them for common sense advice she can trust.
Home Range Ltd has been particularly profitable since the global financial crash beginning in
2007. It appears.
How is Vietas contribution to mathematics distinctly modern in spi.pdf
How is Vieta\'s contribution to mathematics distinctly modern in spirit?
Solution
At the end of 16th century, mathematics was placed under the dual aegis of the Greeks, from
whom they borrowed the tools of geometry, and the Arabs, who provided procedures for the
resolution. At the time of Vieta, algebra therefore oscillated between arithmetic, which gave the
appearance of a list of rules, and geometry which seemed more rigorous. Meanwhile, Italian
mathematicians Luca Pacioli, Scipione del Ferro, Niccolò Fontana Tartaglia, Ludovico Ferrari,
and especially Raphael Bombelli (1560) all developed techniques for solving equations of the
third degree, which heralded a new era.
On the other hand, the German school of the Coss, the Welsh mathematician Robert Recorde
(1550) and the Dutchman Simon Stevin (1581) brought an early algebraic notation, the use of
decimals and exponents. However, complex numbers remained at best a philosophical way of
thinking and Descartes, almost a century after their invention, used them as imaginary numbers.
Only positive solutions were considered and using geometrical proof was common.
The task of the mathematicians was in fact twofold. It was necessary to produce algebra in a
more geometrical way, i.e., to give it a rigorous foundation; and on the other hand, it was
necessary to give geometry a more algebraic sense, allowing the analytical calculation in the
plane. Vieta and Descartes solved this dual task in a double revolution. Firstly, Vieta gave
algebra a foundation as strong as in geometry. He then ended the algebra of procedures (al-Jabr
and Muqabala), creating the first symbolic algebra. In doing so, he did not hesitate to say that
with this new algebra, all problems could be solved (nullum non problema solvere).
In his dedication of the Isagoge to Catherine de Parthenay, Vieta wrote, \"These things which are
new are wont in the beginning to be set forth rudely and formlessly and must then be polished
and perfected in succeeding centuries. Behold, the art which I present is new, but in truth so old,
so spoiled and defiled by the barbarians, that I considered it necessary, in order to introduce an
entirely new form into it, to think out and publish a new vocabulary, having gotten rid of all its
pseudo-technical terms...\"
Vieta did not know \"multiplied\" notation (given by William Oughtred in 1631) or the symbol
of equality, =, an absence which is more striking because Robert Recorde had used the present
symbol for this purpose since 1557 and Guilielmus Xylander had used parallel vertical lines
since 1575.
Vieta had neither much time, nor students able to brilliantly illustrate his method. He took years
in publishing his work, (he was very meticulous) and most importantly, he made a very specific
choice to separate the unknown variables, using consonants for parameters and vowels for
unknowns. In this notation he perhaps followed some older contemporaries, such as Petrus
Ramus, who designated the points in geomet.
Fungal associations with plants Examine a type of fungal association .pdf
Fungal associations with plants Examine a type of fungal association by completing each
sentence. Then, arrange the sentences in a logical paragraph order.
Solution
1. NUTRIENT
The plant provides nutrients to the fungi as fungi doesnot has its green pigment called
Chlorophyll,which helps the plant in photosynthesis.As it doesnot has the chlorophyll,the plant
helps it by providing the nutrients basically the carbon amount so that it will be live and help the
plant in providing lots of nutrients and minerals.
2. MINERAL AND NUTRIENTS
The fungi has different roles in the plant system. The fungi makes a symbiotic relationship with
the plant.By the help of Mycorrhizae ,the plant obtain phosphate and other minerals like Zinc
and Copper . Hyphae are the long extensions which can grow into small soil pores and can allow
the Phosphorous nutrients to the plant body.
3. MYCORRHIZAE, ROOTS
The soil consists of lot of nutrients and minerals. The plant itself cannot intake all the
nutrients,so it takes help of certain symbiotic relationship. By this relationship the plant obtains
Phosphate and other minerals by the help of fungus. While the fungus obtains the sugars from
the roots of the plant.
4. HYPHAE
Hyphae is a lonf,branching, filamentous structure of a fungus that is main mode of vegetative
growth. The hyphae helps the plant in distributiing and absorbing the nutrients and minerals from
the soil..
For each problem in this homework, your assignment is to determine .pdf
For each problem in this homework, your assignment is to determine modes of inheritance for a
set of phenotypes each of which is determined by genotypes at two independently assorting loci.
The data come from an alien species of fly whose genetics is identical to that of earth flies. As
part of your assignment, you must confirm your hypothesized modes of inheritance by applying
a chi-square analysis. NOTE ABOUT DEGREES OF FREEDOM: In situations where the
observed counts and the expected counts for a category are both 0, that category does not figure
into the chi-square calculation, so that the degrees of freedom is reduced by one in such
situations. The possible modes of inheritance are: (i) Autosomal recessive; (ii) Homozygous
lethal (dominant); (iii) Sex-linked recessive. The Sex-linked recessive mode can only occur in
females parents. Counts of phenotype pairs for the F1 generation give you clues as to the mode
of inheritance of each phenotype. Good luck! ONE PHENOTYPE PER PARENT, TWO
PHENOTYPES TOTAL For the Tratagorian Kumbus Fly, that lives on the planet Zeon, parental
matings occur only among flies in which each parent has a disease phenotype at one locus and is
wild-type at the other locus. Problem 4. A female with Flonasean eye color mates with a male
with Muppetrus eye shape, producing the following counts: F1 Generation Males WT-WT
Disease-WT WT-Disease Disease-Disease 0 242 0 247 Females WT-
WT Disease-WT WT-Disease Disease-Disease 239 0 253 0
Solution
It is sex linked recessive mode of inheritance due to which males are not affected and females
are affected..
For each story problem, identofy the opertaion and the interpretatio.pdf
For each story problem, identofy the opertaion and the interpretation of the operationg. Then,
consider the context of each story problem and identify what, if anything, might be problematic
about the context. Pick three of the story problems that you hae idenitfied as having a
problematic context and rewrite each of the story problems.
-Jake, Jenn, and Juile was triplets. For their birthday, their parents bought them 7 presents each.
How many presents did their parents buy all together?
-Terrence scored 27 points in a one-on-one basketball game. Devon scored 34. How many points
did Devon win by?
-Janice gets $40 a week for allowance If she save up for her money for the next 12 weeks, how
much more will she have in all?
-Ramon and Julio got detention after school for 90 minutes every day for a week. How much
time will they spend in detention all together?
Solution
-Jake, Jenn, and Juile was triplets. For their birthday, their parents bought them 7 presents each
parents buy all together 7+7+7 =21 presents
-Terrence scored 27 points in a one-on-one basketball game. Devon scored 34.
Devon win by 34-27 =7 points
-Janice gets $40 a week for allowance If she save up for her money for the next 12 weeks,
she have in all =40*12 =480$
-Ramon and Julio got detention after school for 90 minutes every day for a week.
week has 7 days
time they will spend in detention all together=90*7=630 minutes.
Explain a. casting b. overloading c. sentinel d. echoSolution.pdf
Explain: a. casting b. overloading c. sentinel d. echo
Solution
Casting-
Changing the datatype of 1 variable to another datatype is called casting. Each program has its
own rule.
eg: int can be converted to double.
overloading-
Multiple function when used with same name but have different signature is called overloading
in programs.
sentinel-
It is the flag value that is used in loop like for loop or while which is used for the termination of
loops or recursive algorithm.
eg flag=1
while(flag<5)
{
.....
}
echo-
It is the keyword that is used to print the message or variable in the PHP language
echo \"Hi\";
It will print Hi on screen..
Find all the square roots of the complex number 14i. Write the squar.pdf
Find all the square roots of the complex number 14i. Write the square roots in polar (re^i) form,
with the smaller angle first. Give angles in degress
Solution
this is the polar form r(cos()+isin())
we compare it with 1-4i
rcos =1
rsin=-4
squaring and adding them
r^2(cos^2 +sin^2) = 1 +(-4)^2
r^2 = 1+16
r= sqrt(17) =4.1
tan = -4/1
= arc tan(-4)
= -76 degree
square root of r(cos()+isin()) is ±r(cos(/2)+isin(/2)).
square root form is ±(4.1) (cos(-76/2) + isin(-76/2)
±4.1(cos38 -isin38)
in polar form 4.1 e^(-38i).
Describe the flow of oxygenated blood from the placenta to the fe.pdf
Describe the flow of oxygenated blood from the placenta to the fetus\' systemic circulation and
deoxygenated blood from the fetus to the placenta. Be sure to include all five structures that
allow for fetal circulation to take place effectively.
Solution
Placenta receives blood from two systems: a uteroplacental system that delivers maternal blood
to the placenta; and the fetoplacental system that receives from the fetus.
The umbilical vein carries blood from the placenta to the fetus. A majority of it enters liver and
1/3rd is diverted to ductus venosus and carried to inferior vena cava. In liver proper, the blood
enters from its inferior border. A branch of the umbilical vein joins the portal vein. The blood is
then supplied to the right atrium of the heart. The heart has an opening connecting left and right
atrium called foramen ovale. Due to this opening, the blood bypasses the pulmonary circulation
and is directly circulated to the body of the fetus, via the aorta. Some of the blood moves to the
placenta through iliac and then umbilical arteries into the mother\'s blood circulation for the
exchange of gasses and removal of waste products.
A part of the blood that enters the atrium of the fetus enters left ventricle and is pumped to the
pulmonary artery The pulmonary artery of the fetus is connected to the aorta via ductus atriosus
and thus the blood enters aorta and escapes from entering the lungs. The blood in aorta takes the
same route as described earlier..
describe two different forms of bindingSolutionTwo forms Co.pdf
describe two different forms of binding
Solution
Two forms :
Compile time Binding : thus is done at compile time where the symbols are resolved i.e the type
is assigned during compile time.also called early or static binding.
Runtime binding where the functions, variables methods and properties are assiged type during
run time. also called as late or dynamic binding.
Examples:
Compile time - method overloading
Run time - method overriding.
Connect onnect mheducation.co.pdf
Connect onnect mheducation.com/flow/connect.html?isReg true&returnuri;
https%3A%2F%2Fconnect.mheducation com%2Fpa C a Search uiz (Lecture 4) Taxonomy is the
branch of systemic biology that studies the evolutionary relatedness among groups of organisms.
Answer True False Help
Solution
2. False
Taxonomy is the science of naming, describing and classifying living organisms
3. Presence of a jaw
4. True.
Consider the following interrupting system. The active-edge inputs o.pdf
Consider the following interrupting system. The active-edge inputs on PJ7 and PP7 can occur at
any time, including at the same time. The object is to count the number of each type of interrupt.
How would you best describe the usage of Count in this system? a) This is a perfectly
appropriate usage of Count, because there are two permanently allocated variables with private
scope, such that each variable counts the number of interrupts for each ISR. b) There is a critical
section bug because of the read/modify/write access to a shared global. c) Because both ISRs
share the same Count, the system can not distinguish between a PJ7 and a PP7 interrupt. d)
Count is initialized each interrupt, so its value is not the total number. e) The acknowledge
statements in the two ISRs are not friendly because they affect all 8 bits of the flag register. f)
None of the above is true. void interrupt 24 KeyHanJ (void) ( unsigned short static Count-0; PIFJ
= 0x80; // acknowledge Countt count the number of PU7 edges void interrupt 56 KeyHanP
(void) ( unsigned short static count=0; PIFP = 0x80; // acknowledge Count++count the number
of PP7 edges
Solution
Looking around, we find ourselves to be surrounded by various types of embedded systems. Be
it a digital camera or a mobile phone or a washing machine, all of them has some kind of
processor functioning inside it. Associated with each processor is the embedded software. If
hardware forms the body of an embedded system, embedded processor acts as the brain, and
embedded software forms its soul. It is the embedded software which primarily governs the
functioning of embedded systems.
During infancy years of microprocessor based systems, programs were developed using
assemblers and fused into the EPROMs. There used to be no mechanism to find what the
program was doing. LEDs, switches, etc. were used to check correct execution of the program.
Some ‘very fortunate’ developers had In-circuit Simulators (ICEs), but they were too costly and
were not quite reliable as well.
As time progressed, use of microprocessor-specific assembly-only as the programming language
reduced and embedded systems moved onto C as the embedded programming language of
choice. C is the most widely used programming language for embedded processors/controllers.
Assembly is also used but mainly to implement those portions of the code where very high
timing accuracy, code size efficiency, etc. are prime requirements.
Initially C was developed by Kernighan and Ritchie to fit into the space of 8K and to write
(portable) operating systems. Originally it was implemented on UNIX operating systems. As it
was intended for operating systems development, it can manipulate memory addresses. Also, it
allowed programmers to write very compact codes. This has given it the reputation as the
language of choice for hackers too.
As assembly language programs are specific to a processor, assembly language didn’t offer
portability across systems. To overcome this disadv.
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