The weak D phenotype is thought to arise from several mechanisms. .pdf
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The weak D phenotype is thought to arise from several mechanisms. Which one of the following statements is false? C gene is in trans position to D gene, this results in a weakening influence on the D gene, examples CDe/Cde and cDe/Cde. Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in caucasians. Also occurs in the Ro genotype in certain racial groups more commonly. The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these pieces is missing. The red cells are said to have a partial D. D deletion genes gives rise to weakened expression of D antigen. Rh null cells lack: MNSs antigens. Lewis antigens. Normal oxygen-carrying capacity. Rh antigens. There are two closely related and closely linked genes. Enzyme techniques, in conjunction with an IgM anti-D. Increasing the time for the incubation of the saline phase. Performance of an IAT using an IgG anti-D. Don\'t need to do it in a blood bank. C gene is in trans position to D gene, this results in a weakening influence on the D gene, examples CDe/Cde and cDe/Cde. Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in caucasians. Also occurs in the Ro genotype in certain racial groups more commonly. The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these pieces is missing. The red cells are said to have a partial D. D deletion genes gives rise to weakened expression of D antigen. Solution The weak D phenotype is thought to arise from several mechanisms. Which one of the following statements is false? D deletion genes gives rise to weakened expression of D antigen. Rh null cells lack: Rh antigens. There are two closely related and closely linked genes. Enzyme techniques, in conjunction with an IgM anti-D. Increasing the time for the incubation of the saline phase. Performance of an IAT using an IgG anti-D. Don\'t need to do it in a blood bank. Last one is an incomplete question D deletion genes gives rise to weakened expression of D antigen..
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In a comparison of ectotherms and endotherms, which of the following.pdf
In a comparison of ectotherms and endotherms, which of the following are correct?
1. Endotherms are regulators while ectotherms are conformers
2. Endotherms have higher energetic demands compared to ectotherms.
3. Endotherms spend more energy on internal temperature regulation compared to ectotherms.
4. Endothermy is more common among animal phyla than ectothermy.
5. At the cellular level, endotherms cells always produce more ATP and more heat.
Solution
1.correct
Regulator
A regulator is also known as an endotherm. A regulator uses internal control mechanisms to
maintain the optimum temperature needed for its body to function. Mammals and birds are
examples of regulators. They are often described as warm blooded because they maintain a
constant body temperature. For example, if the climate around a regulator gets hotter, the animal
can adjust its metabolism to lessen the need for heat production.
Conformer (or ectotherm)
A conformer is also called an ectotherm. A conformer cannot regulate its own internal
temperature so it must adapt its behavior, physically moving itself to environments with
optimum temperatures. Reptiles, amphibians and insects are all conformers and are often referred
to as cold blooded. For example, a lizard may lie on a rock in the sun to raise its body
temperature, as it cannot regulate it internally.
2.not correct
Endotherms also called warm blooded animals have the ability to regulate their body temperature
by themselves.
If the endotherms are in a cold place their body increases the metabolism and generates more
heat. This will compensate for the cold outside.
In a hot region the body metabolism is tuned down (this is not as efficient as the adaptation to
cold region - as the metabolism can\'t be turned down below a certain limit - this will result in
death) a little and heat generation is somewhat reduced. The second and more efficient
mechanism is the heat dissipation by sweating and through skin (directly).
Ectotherms on the other hand, do not have efficient temperature regulation systems.
In warm regions when the temperature is high, the ectotherms seek shade and low-temperature
areas to prevent over-heating
In cool regions they seek sunlight and expose themselves to the sunlight for warmth. For the
same reason, they can also search out and stay near fire pits, campfires etc...
So when we calculate the energy demands, the ectotherms need more energy as they regulate
their own metabolisms to match the need.
3.correct
previously explaine
4.not correct,ectother
ectothermy is more common then endothermy
5.correct.
Implement Breadth-First Search with a QueueSolution Program .pdf
Implement Breadth-First Search with a Queue
Solution
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
for(i = adj[s].begin(); i != adj[s].end(); ++i)
{
if(!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}
// Driver program to test methods of graph class
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
cout << \"Following is Breadth First Traversal \"
<< \"(starting from vertex 2) \ \";
g.BFS(2);
return 0;
}
Output:
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
fo.
Immunology what are TdT and AID, and what are their roles in B c.pdf
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Solution
Answer-TdT stands for terminal deoxynucleotidel transferase.i t is a special kind of DNA
polymerase expressed immature pre B and pre T lymphocytes.Unlike other DNA polymerase
TdT has a unusual function to synthesize DNA without using template DNA.
AID-AID stands for activation induced cytidine deaminase.i T is a 24kDa enzyme in human
which is responsible for antibody diversity.d ye to its deamination activity it deaminase it causes
mutation in DNA by deamination of cytosine to uracil and thus change base pairs G-C to U-Aor
T-A due to these mutation in germinal center of B lymphocytes antibody diversity produces.
Role of TdT and AID in B cells-Bcells are immune cell that produces antibody with the binding
sites for antigen. Different types of antibodies produced by B cell which give immunity to
different type of antigens.These antibody diversity occure due to VDJ recombination .In
antibody producing pre B cells rearrangement of these variable(v), diversity (D), joining (J)gene
segments allow B cells to generate antibody with different antigen binding sites with antibody
specificity.For breaking,arranging and rejoining these gene segments district classes of enzymes
like nuclease ,ligase and polymerase work.S specific kind of polymerase used for these gene
recombination known as TdT.
RAG enzyme cut- the double standard DNA which contains genes .and the hairpin structure left
is open by Artemis complex which has endonuclease activity.once Artemis complex open
hairpin and add palindromic sequence in DNA cut end TdT start working on 3\'OH end of DNA
and add nucleotide to single stranded DNA in 5\'to3\' direction without using template.single
stranded DNA formed at both the ends of non homologous segment joining done by Watson and
Crick base pairing and unpaired bases are removed by exonuclease ,gap is filled by template
dependent polymerase and Nick is sealed by ligase.in this way due to the recombination of
antibody diversity producing gene pre B cells produces diffrent type of antibody specific to
different type of antigen.
AID produces mutation in B cells during germinal center development by deamination of
cytosine that produces antibody diversity.
CD-R are the part of variable chain in immunoglobulin generated by B cells and T cell
receptor.tdt and AID helps to crate these sites on antibody and create antibody diversity..
IDS in the Cloud Please respond to the following(a) Explain w.pdf
\"IDS in the Cloud\" Please respond to the following:
(a) Explain whether or not you believe technologies such as IDS are still relevant and useful as
there is a push toward SaaS and cloud-based solutions.
(b) Discuss from your perspective how cloud-based services change incident response, for better
or worse, and determine what you believe to be the greatest preparedness concern with cloud-
based services.
PLEASE NO COPY AND PASTING
Solution
(a)
IDS - Intrusion Detection systems
To prevent threat over once computer security firewall and encryption techniques were invented.
Although, the intruders could still hack one\'s system. To prevent this, Intrusion detection
systems were introduced which provide the following advantages:
1.Continuously monitors and analyzes the activities of the user and system.
2.Uses pattern recognition for recognizing typical kinds of attacks.
3.Tracks user policy violations.
3. Recognizes abnormal activities through pattern recognition.
SaaS: Software as a Service
1.SaaS is a model for Software licensing and delivery which is hosted on internet on subscription
basis and made available for the internet users.
2. It is one of the categories of cloud computing, where in instead of simply installing and
maintaining software you keep it over the internet which is easily maintainable from complex
software management.
Intrusion detection systems are still relevant to protect one\'s private data although Cloud
computing exists.
(b) Security and Incident response is a major subject which many vendors try to overcome.
At high level, the process defines six stages of incident response :
Preparation,identification,containment,eradication,recovery,lessons learned.
Five challenges or risks or concerns faced with cloud computing:
1.Privacy and Security.
2.Cloud transition.
3.Data access
4.Service quality
5.Data accessibility..
Gandalf the Grey started in the Forest of Mirkwood at a point with co.pdf
Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (-3, -1) and
arrived in the Iron Hills at the point with coordinates (-1,3). If he began walking in the direction
of the vector v = 5i + lj and changes direction only once, when he turns at a right angle, what are
the coordinates of the point where he makes the turn.
Solution
Geometrically, the vector 5i + 1j corresponds to a slope of 1/5.
So, at the right angle where Gandalf changes direction, the slope becomes -5.
Let\'s write out the equations.
Point (-3, -1), slope 1/5 => y + 1 = (1/5)(x + 3) or y = (1/5)(x + 3) - 1 or y = (1/5)(x - 2)
Point (-1, 3), slope - 5 => y - 3 = (-5)(x + 1) or y = - 5x - 2.
Now, we find the intersecting point of these two curves
(1/5)(x - 2) = -5x - 2
==> x = - 4/13
Substituting this into either equation yields y = - 6/13.
So, the desired point is ( - 4/13, - 6/13)..
Explain exploratory factor analysis and give real life examplesS.pdf
Explain exploratory factor analysis and give real life examples
Solution
In multivariate statistics, exploratory factor analysis (EFA) is a statistical method used to
uncover the underlying structure of a relatively large set of variables. EFA is a technique within
factor analysis whose overarching goal is to identify the underlying relationships between
measured variables.[1] It is commonly used by researchers when developing a scale (a scale is a
collection of questions used to measure a particular research topic) and serves to identify a set of
latent constructs underlying a battery of measured variables.[2] It should be used when the
researcher has no a priori hypothesis about factors or patterns of measured variables.[3]Measured
variables are any one of several attributes of people that may be observed and measured. An
example of a measured variable would be the physical height of a human being. Researchers
must carefully consider the number of measured variables to include in the analysis.[2] EFA
procedures are more accurate when each factor is represented by multiple measured variables in
the analysis.
Following is the set of EFA examples,
. • 4.1: Exploratory factor analysis with continuous factor indicators
• 4.2: Exploratory factor analysis with categorical factor indicators
• 4.3: Exploratory factor analysis with continuous, censored, categorical, and count factor
indicators*
• 4.4: Exploratory factor mixture analysis with continuous latent class indicators
• 4.5: Two-level exploratory factor analysis with continuous factor indicators
• 4.6: Two-level exploratory factor analysis with both individual- and cluster-level factor
indicators
• 4.7: Bi-factor exploratory factor analysis with continuous factor indicators
* Example uses numerical integration in the estimation of the model. This can be
computationally demanding depending on the size of the problem..
Drag the labels to the correct category to demonstrate your understan.pdf
Drag the labels to the correct category to demonstrate your understanding of cell walls in
different types of microbes. Protozoans Helminths Bacteria Fungi
Solution
Protozoans- These are classified under kingdom Protist.
Helminths-These are classified under kingdom animalia.
Fungi-These are classified under kingdom fungi.
Bacteria- Prokaryotes..
Do you agree that the Berg Letter, research moratorium, and Asilomar.pdf
Do you agree that the Berg Letter, research moratorium, and Asilomar Conference were
necessary before further recombinant DNA research was conducted? Why or why not?
Solution
Paul Berg, an American biochemist and professor emeritus at Stanford University witnessed
firsthand the history of recombinant DNA research and regulation, having been in the forefront
of both movements since he was a young man. He gained early recognition and influence when
he delineated the key steps in which DNA produces proteins.
In the mid-1970s, the National Academy of Sciences asked Berg to explore the safety of
recombinant DNA technology. He responded with the historic \"Berg letter,\" calling for a
moratorium on recombinant DNA research until safety issues could be addressed. He was one of
the key organizers of the international forum on recombinant DNA technology, the Asilomar
Conference. The ensuing dialogue resulted in the National Institutes of Health guidelines
published a year later, a milestone of responsible self-regulation in science.
The Asilomar Conference on Recombinant DNA was an influential conference organized by
Paul Berg to discuss the potential biohazards and regulation of biotechnology, held in February
1975 at a conference center at Asilomar State Beach.A group of about 140 professionals
(primarily biologists, but also including lawyers and physicians) participated in the conference to
draw up voluntary guidelines to ensure the safety of recombinant DNA technology. The
conference also placed scientific research more into the public domain, and can be seen as
applying a version of the precautionary principle.
The effects of these guidelines are still being felt through the biotechnology industry and the
participation of the general public in scientific discourse. Due to potential safety hazards,
scientists worldwide had halted experiments using recombinant DNA technology, which entailed
combining DNAs from different organisms. After the establishment of the guidelines during the
conference, scientists continued with their research, which increased fundamental knowledge
about biology and the public’s interest in biomedical research.
During the conference, the principles guiding the recommendations for how to conduct
experiments using this technology safely were established. The first principle for dealing with
potential risks was that containment should be made an essential consideration in the
experimental design. A second principle was that the effectiveness of the containment should
match the estimated risk as closely as possible.
The conference also suggested the use of biological barriers to limit the spread of recombinant
DNA. Such biological barriers included fastidious bacterial hosts that were unable to survive in
natural environments. Other barriers were nontransmissible and equally fastidious vectors
(plasmids, bacteriophages, or other viruses) that were able to grow in only specified hosts.
In addition to biological barriers, the co.
Brief Concepts and Questions EXAM 2 Chapter 8 DNA RNA Protein What i.pdf
Brief Concepts and Questions EXAM 2 Chapter 8: DNA RNA Protein What is DNA? a
phosphate Structure of DNA: Building blocks are called nucleotides Each nucleotide is
composed of three br uithofenas bee. What makes DNA so special? Provide 4 reasons, below
DNA DNA (Replication): Where does DNA replication take place? When does DNA replication
take place? Explain steps involved in DNA replication: DNA RNA Protein (Gene Expression)
Involves 2 processes: 1. Transcription 2. Translation Explain the Synthesis of Proteins (Gene
Expression): o DNA RNA Protein What is RNA? What is \"codon What is \"anticodon\" What is
a protein molecule? DNA mutation; Change in nucleotide bases of DNA Duplex Point mutation
Frame shift mutation
Solution
Question
Answer
Where does DNA replication take place:
It takes place in the nucleus in case of eukaryotic cells and in the cytoplasm in case of
prokaryotic cells
When does DNA replication take place:
DNA replication occurs during the S-phase during cell cycle, so that cell can make an extra copy
of genetic material.
Explain steps involved in DNA replication:
Initiation: During initiation, the proteins will bind to the origin of replication; helicase unwinds
the DNA helix which results in the formation of two replication forks.
Elongation: A RNA primer sequence will be added to this the DNA pol III will add the
nucleotides in 5’ to 3’ direction and chain will elongate.
Termination: In case of bacteria, termination of replication occurs whenever two replication
forks meet each other from the opposite end of the parental chromosome.
Transcription
Gene expression first step is transcription, here a particular segment of DNA will be copied into
RNA with the help of the enzyme RNA polymerase
Translation
Translation is the final step of the gene expression. Here mRNA will be used to synthesize the
polypeptide chain. The information present in the mRNA in the form of codon will code for the
amino acids needed for polypeptide chain synthesis.
What is RNA?
RNA is ribonucleic acid and is found in all living cells. It acts as the messenger carrying
instructions from DNA for the synthesis of proteins.
Few viruses will have RNA as their genetic material.
What is codon?
Codon is a sequence of three nucleotides and they together form a unit of genetic code in either
DNA or RNA.
What is anticodon?
It is found on tRNA and it is a sequence of three nucleotides which forms a genetic code on
tRNA, and these anticodon is complementary to the codons found on messenger RNA.
What is a protein molecule?
During translation, when amino acids are added in a sequential manner, the condensation of
amino acids will form a peptide bond in between them and finally forms a polypeptide chain. It
is the DNA through mRNA directs the protein synthesis.
Point mutation
In point mutation, only one or very few nucleotides will be affected or mutated in a gene
sequence.
Frame shift mutation
Either insertions or deletion can result in frame shift mutation, due to th.
A population has two alleles for a particular gene, A and a, and is .pdf
A population has two alleles for a particular gene, A and a, and is in Hardy-Weinberg
equilibrium. The frequency of allele a is 0.2. What is the frequency of individuals with AA
genotype?
0.2
0.4
0.64
0.04
0.8
Solution
C). 0.64
Hardy and Weinberg mathematically proved that in a population, all dominant and recessive
alleles comprise all alleles for that gene.
This was mathematically represented as p+ q = 1.0
Where,
p = frequency of dominant alleles
q = frequency of recessive alleles.
Given that, the frequency of a = 0.2
As the population is in H-W equilibrium, the frequency of A = 1-0.2 = 0.8
The frequency of AA genotype = 0.8*0.8 = 0.64.
A 12-year-old male complained of strange odors before loss of consci.pdf
A 12-year-old male complained of strange odors before loss of consciousness and a major tonic-
clonic seizure. During the seizure, his parents rushed him to the emergency room of a local
hospital. On arrival at the hospital, he appeared to be asleep. Studies at the hospital showed
routine laboratory work within normal limits (WNL), lumbar puncture (CSF) was WNL, no
evidence of skull fracture on x-ray study was revealed, and an electroencephalograph (EEG)
showed no abnor malities. How would you interpret the episode and findings?
Solution
Charecterized by sudden loss of consciousness ,chaotic muscle movement and tone, and apnea
Tonic phase:Bilateral muscle rigidity
Clonic phase: muslce contraction and relaxation lasting 1 to 3 minutes
Signs:
Tachycardia, hyerpventilation,sweating, and intense salivation
Last 3-5 minutes
postictal state(5 to 20 mintues)follows
-gradual return to consciousness.
2.Which of the following statements about evolution is falseNonrand.pdf
2.Which of the following statements about evolution is false?Nonrandom mating can change the
genetic makeup of populations over time.Genetic drift can change the genetic makeup of
populations over time.Natural selection is one of several mechanisms of evolution.Populations,
not individuals, evolve.Evolution is goal-directed.2.Which of the following statements about
evolution is false?Nonrandom mating can change the genetic makeup of populations over
time.Genetic drift can change the genetic makeup of populations over time.Natural selection is
one of several mechanisms of evolution.Populations, not individuals, evolve.Evolution is goal-
directed.
Solution
Note:.
can someone explain in a easy way chromosome walking I know its w.pdf
can someone explain in a easy way chromosome walking? I know it\'s with overlapping
fragments and a gen nearby the gen you want to find, but I don\'t clearly understand it, I have
images in my book, but an image with steps and explanation would be nice,
Solution
Chromosome Walking:
Chromosome walking is positional cloning of gene sequences or upstream promoter sequences.
Eukaryotic DNA contains a lot of junk DNA between the genes. The technique relies upon
screening of gDNA library for overlapping fragments in a sequential manner to tap the upstream
or downstream region.
Procedure:
1. Isolation of Genomic DNA: First of all isolate genomic DNA (gDNA) of the test organism
and then partially digest it with some restriction enzyme, say Hind III that would yield unequal
random fragments.
2. Cloning of the Digested into E.coli: Clone the digested gDNA into a cloning vector (between
the Hind III sites in this case) to prepare a gDNA library of the organism. Each colony of the
library would have only one clone.
3. Preparation of marker probe: Prepare a radioactively labelled probe from the neighboring gene
(say A), from where you want to start walking.
4. Colony Hybridization: Hybridize the probe with your gDNA library. The probe would show
the signal in the colonies carrying DNA in and around your gene.
5. Sequencing: Pick the colonies showing the signal and isolate the plasmid DNA and sequence
each using the plasmid-specific primers. You would get only one sequence per bacterial colony
of your library.
The sequence, thus obtained must be analyzed. Select longer sequences towards the direction of
your walk.
6. Preparation of the probe: Prepare new probes from the region towards your target gene.
7. Repeat the steps 4-6 till you get the clone and sequence of your target gene..
9. When a bacteriophage inserts its DNA into a bacterial chromosome,.pdf
9. When a bacteriophage inserts its DNA into a bacterial chromosome, the inserted
bacteriophage DNA is called _____
(a) prophage
(b) lytogenic
(c) virulent
Solution
a.Prophage.
The genetic material of a bacteriophage ,incorporated in to the genome of bacterium or host ,now
it is ready to prouce phages if specifically activated.
This stage of Bacteriophage life cycle called as Prophage..
Write an informal negation for each of the following statements. Be .pdf
Write an informal negation for each of the following statements. Be careful to avoid negations
that are ambiguous. a. All dogs are friendly. b. All people are happy. c. Some suspicions were
substantiated. d. Some estimates are accurate.
Solution
(a)
these is a dog which is not friendly
(b)
Some people are not happy
(c)
these was a ssuspicions which was not substantiated
(d)
these is some estimates which are wrong.
Write a C++ program which generates and displays a random walk acros.pdf
Write a C++ program which generates and displays a random walk across a grid. A 2-
dimensional array of characters will be used for this grid. The cells visited by the walk will be
labeled by capital letters with A being the first cell, B the cell moved to from A, and so on,
ending either when Z is reached or when it is not possible to move to another cell.
The following libraries, constant definition, and type definition are needed in the program:
#include
#include // for srand and rand
#include // for time
using namespace std;
const int SIZE = 10;
typedef char Grid[SIZE][SIZE]
The grid must be set to have all its elements be the character ’.’ before the walk is begun.
The program must contain the following function prototypes, the main function, and the function
definitions for all the other functions:
// Function prototypes
bool can_move_up(int i, int j, Grid walk);
bool can_move_down(int i, int j, Grid walk);
bool can_move_left(int i, int j, Grid walk);
bool can_move_right(int i, int j, Grid walk);
void init_array(Grid walk);
void generate_random_walk(Grid walk);
void print_array(Grid walk);
int main(void)
{
Grid walk; // the grid in which the random walk occurs
srand((unsigned) time(NULL));
init_array(walk);
generate_random_walk(walk);
print_array(walk);
return 0;
}
The four functions can_move _* must return whether or not it is possible to move in the
particular direction in the grid. For example, a function call like if (can_move_up(i, j, walk))
could be used to determine whether or not it is possible to move up. The rule for a move being
possible is that if it doesn’t go outside of the grid and isn’t already part of the walk, then that
move is possible. The function init_array must set all characters in the grid to ’.’
The function generate_random_walk must modify the cells of grid by changing ones that are
part of the walk to letters.
The function print_array must print the resulting grid in the form shown in the sample run
below.
In the function generate_random_walk, at each step of the walk, randomly select one of the 4
directions and try to move in that direction. If that move isn’t available, try the other moves in
turn until one is possible. If all 4 directions are blocked, the walk must terminate, and the
program must continue and print out the walk. To select one of the 4 directions randomly, use
the expression: rand() % 4
The function rand returns an integer, and finding the remainder of this integer divided by 4
results in getting one of the numbers 0, 1, 2, or 3. Use these 4 numbers to represent the 4
directions up, down, left and right.
The call to the function srand from the main function seeds the pseudo-random number
generator used by rand so that different sequences of numbers will be generated by calls to rand
each time the program is run.
Solution
#include //input output functions in C++
#include //rand() and srand() functions
#include //time() function for creating seed
#include
#include // for srand and rand
#inclu.
true or false1- The grouped frequency table should ideally have.pdf
true or false???
1- The grouped frequency table should ideally have 10 class intervals
2- The width of the intervals should be half of the number of scores
3- Each score can be represented in one and only one interval
4- The intervals should end with the highest number in the sample
Solution
1.True
2.False
3.True
4.False.
Answer the questions that follow the VHDL code given below. LIBRARY .pdf
Answer the questions that follow the VHDL code given below. LIBRARY ieee; USE
ieee.std_logic_1164.ALL; ENTITY ex1_IS PORT (n, cp, n rd: IN st_logic; q: BUFFER integer
RANGE 0 TO 15); END ex1; ARCHITECTURE are OF ex1 IS BEGIN PROCESS (n_cp,
n_rd) BEGIN IF (n_rd = \'0\') THEN q
Solution
a)n_rd- used to reset the counter
n_cp-used as clock to the counter
b)4 binary bits are there in q;
Q stores 0 t0 15 number so
log2(16)=4.
The coding region of a BACTERIAL gene has the following sequence A.pdf
The coding region of a BACTERIAL gene has the following sequence:
\"AGAGCCAACATGTCGAGCATATGCTGATATACG\' Draw and label the DNA (including
ALLEABIS of the gene, sequence information) The info you need is in the sides. Label the
template/coding strands Label 5\' and 3\' ends of each DNA strand Label where RNA
polymerase binds \"Transcribe\" this gene into mRNA, as shown in Bb video. Write the mRNA
nucleotide sequence, starting at the correct nucleotide, starting at the correct nucleotide Show
the 5\' and 3\' ends of the RNA. Use the codon chart to \"translate\" the mRNA into a protein,
and write the correct amino acid sequence of the protein. Only translate the part of the gene that
a cell would translate. Label the N-terminus and C-terminus of the protein. If the sequence of
the gene were changed to: AGACCCAACATGTCGGACATATGCTGATATACG what would
happen to the properties of the protein? Explain why write out the entire process of bacterial
transcription step by step in the correct order, in detail.
Solution
4. Process of Bacterial Transcription:
Bacterial transcription is the process in which mRNA transcripts of genetic material in bacteria
are produced; it is further translated to produce proteins. Unlike in eukaryotes, bacterial
transcription and translation can occur simultaneously in the cytoplasm.
RNA polymerases are enzymes that transcribe DNA into RNA. Using a DNA template, RNA
polymerase builds a new RNA molecule through base pairing. For instance, if there is a G in the
DNA template, RNA polymerase will add a C to the new, growing RNA strand.
Three steps are involved in bacterial transcription:
Initiation: To begin transcribing a gene, RNA polymerase binds to the DNA of the gene at a
region called the promoter. Basically, the promoter tells the polymerase where to locate on the
DNA and begin transcribing.
Elongation:
Once RNA polymerase is in position at the promoter, the next step of
transcription—elongation—can begin. Basically, elongation is the stage when the RNA strand
gets longer, thanks to the addition of new nucleotides.
During elongation, RNA polymerase \"walks\" along one strand of DNA, known as the template
strand, in the 3\' to 5\' direction. For each nucleotide in the template, RNA polymerase adds a
matching (complementary) RNA nucleotide to the 3\' end of the RNA strand.
Termination:
RNA polymerase will keep transcribing until it gets signals to stop. The process of ending
transcription is called termination, and it happens once the polymerase transcribes a sequence of
DNA known as a terminator.
There are two major termination strategies found in bacteria: Rho-dependent and Rho-
independent.
In Rho-dependent termination, the RNA contains a binding site for a protein called Rho factor.
Rho factor binds to this sequence and starts \"climbing\" up the transcript towards RNA
polymerase.
When it catches up with the polymerase at the transcription bubble, Rho pulls the RNA transcript
and the template DNA strand apart, releasing .
Regarding the ABO blood group system in humans, if an individual is g.pdf
Regarding the ABO blood group system in humans, if an individual is genetically I^B i and yet
expresses the O blood type, which genotype is he or she likely to have? Hh HH hh i^A i i^B i
none of the above are correct.
Solution
Option C is the correct answer. This hh is also known as Bombay blood group. These individuals
do not express H antigen. Thus these individuals are unable to form either of A or B antigen on
their RBCs. Usual test of these individuals for blood group shows them as having O blood group..
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
Delivering Micro-Credentials in Technical and Vocational Education and Training
Explore how micro-credentials are transforming Technical and Vocational Education and Training (TVET) with this comprehensive slide deck. Discover what micro-credentials are, their importance in TVET, the advantages they offer, and the insights from industry experts. Additionally, learn about the top software applications available for creating and managing micro-credentials. This presentation also includes valuable resources and a discussion on the future of these specialised certifications.
For more detailed information on delivering micro-credentials in TVET, visit this https://tvettrainer.com/delivering-micro-credentials-in-tvet/
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In a comparison of ectotherms and endotherms, which of the following.pdf
In a comparison of ectotherms and endotherms, which of the following are correct?
1. Endotherms are regulators while ectotherms are conformers
2. Endotherms have higher energetic demands compared to ectotherms.
3. Endotherms spend more energy on internal temperature regulation compared to ectotherms.
4. Endothermy is more common among animal phyla than ectothermy.
5. At the cellular level, endotherms cells always produce more ATP and more heat.
Solution
1.correct
Regulator
A regulator is also known as an endotherm. A regulator uses internal control mechanisms to
maintain the optimum temperature needed for its body to function. Mammals and birds are
examples of regulators. They are often described as warm blooded because they maintain a
constant body temperature. For example, if the climate around a regulator gets hotter, the animal
can adjust its metabolism to lessen the need for heat production.
Conformer (or ectotherm)
A conformer is also called an ectotherm. A conformer cannot regulate its own internal
temperature so it must adapt its behavior, physically moving itself to environments with
optimum temperatures. Reptiles, amphibians and insects are all conformers and are often referred
to as cold blooded. For example, a lizard may lie on a rock in the sun to raise its body
temperature, as it cannot regulate it internally.
2.not correct
Endotherms also called warm blooded animals have the ability to regulate their body temperature
by themselves.
If the endotherms are in a cold place their body increases the metabolism and generates more
heat. This will compensate for the cold outside.
In a hot region the body metabolism is tuned down (this is not as efficient as the adaptation to
cold region - as the metabolism can\'t be turned down below a certain limit - this will result in
death) a little and heat generation is somewhat reduced. The second and more efficient
mechanism is the heat dissipation by sweating and through skin (directly).
Ectotherms on the other hand, do not have efficient temperature regulation systems.
In warm regions when the temperature is high, the ectotherms seek shade and low-temperature
areas to prevent over-heating
In cool regions they seek sunlight and expose themselves to the sunlight for warmth. For the
same reason, they can also search out and stay near fire pits, campfires etc...
So when we calculate the energy demands, the ectotherms need more energy as they regulate
their own metabolisms to match the need.
3.correct
previously explaine
4.not correct,ectother
ectothermy is more common then endothermy
5.correct.
Implement Breadth-First Search with a QueueSolution Program .pdf
Implement Breadth-First Search with a Queue
Solution
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
for(i = adj[s].begin(); i != adj[s].end(); ++i)
{
if(!visited[*i])
{
visited[*i] = true;
queue.push_back(*i);
}
}
}
}
// Driver program to test methods of graph class
int main()
{
// Create a graph given in the above diagram
Graph g(4);
g.addEdge(0, 1);
g.addEdge(0, 2);
g.addEdge(1, 2);
g.addEdge(2, 0);
g.addEdge(2, 3);
g.addEdge(3, 3);
cout << \"Following is Breadth First Traversal \"
<< \"(starting from vertex 2) \ \";
g.BFS(2);
return 0;
}
Output:
// Program to print BFS traversal from a given source vertex. BFS(int s)
// traverses vertices reachable from s.
#include
#include
using namespace std;
// This class represents a directed graph using adjacency list representation
class Graph
{
int V; // No. of vertices
list *adj; // Pointer to an array containing adjacency lists
public:
Graph(int V); // Constructor
void addEdge(int v, int w); // function to add an edge to graph
void BFS(int s); // prints BFS traversal from a given source s
};
Graph::Graph(int V)
{
this->V = V;
adj = new list[V];
}
void Graph::addEdge(int v, int w)
{
adj[v].push_back(w); // Add w to v’s list.
}
void Graph::BFS(int s)
{
// Mark all the vertices as not visited
bool *visited = new bool[V];
for(int i = 0; i < V; i++)
visited[i] = false;
// Create a queue for BFS
list queue;
// Mark the current node as visited and enqueue it
visited[s] = true;
queue.push_back(s);
// \'i\' will be used to get all adjacent vertices of a vertex
list::iterator i;
while(!queue.empty())
{
// Dequeue a vertex from queue and print it
s = queue.front();
cout << s << \" \";
queue.pop_front();
// Get all adjacent vertices of the dequeued vertex s
// If a adjacent has not been visited, then mark it visited
// and enqueue it
fo.
Immunology what are TdT and AID, and what are their roles in B c.pdf
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Immunology: what are TdT and AID, and what are their roles in B cells? How does this relate to
CDRs?
Solution
Answer-TdT stands for terminal deoxynucleotidel transferase.i t is a special kind of DNA
polymerase expressed immature pre B and pre T lymphocytes.Unlike other DNA polymerase
TdT has a unusual function to synthesize DNA without using template DNA.
AID-AID stands for activation induced cytidine deaminase.i T is a 24kDa enzyme in human
which is responsible for antibody diversity.d ye to its deamination activity it deaminase it causes
mutation in DNA by deamination of cytosine to uracil and thus change base pairs G-C to U-Aor
T-A due to these mutation in germinal center of B lymphocytes antibody diversity produces.
Role of TdT and AID in B cells-Bcells are immune cell that produces antibody with the binding
sites for antigen. Different types of antibodies produced by B cell which give immunity to
different type of antigens.These antibody diversity occure due to VDJ recombination .In
antibody producing pre B cells rearrangement of these variable(v), diversity (D), joining (J)gene
segments allow B cells to generate antibody with different antigen binding sites with antibody
specificity.For breaking,arranging and rejoining these gene segments district classes of enzymes
like nuclease ,ligase and polymerase work.S specific kind of polymerase used for these gene
recombination known as TdT.
RAG enzyme cut- the double standard DNA which contains genes .and the hairpin structure left
is open by Artemis complex which has endonuclease activity.once Artemis complex open
hairpin and add palindromic sequence in DNA cut end TdT start working on 3\'OH end of DNA
and add nucleotide to single stranded DNA in 5\'to3\' direction without using template.single
stranded DNA formed at both the ends of non homologous segment joining done by Watson and
Crick base pairing and unpaired bases are removed by exonuclease ,gap is filled by template
dependent polymerase and Nick is sealed by ligase.in this way due to the recombination of
antibody diversity producing gene pre B cells produces diffrent type of antibody specific to
different type of antigen.
AID produces mutation in B cells during germinal center development by deamination of
cytosine that produces antibody diversity.
CD-R are the part of variable chain in immunoglobulin generated by B cells and T cell
receptor.tdt and AID helps to crate these sites on antibody and create antibody diversity..
IDS in the Cloud Please respond to the following(a) Explain w.pdf
\"IDS in the Cloud\" Please respond to the following:
(a) Explain whether or not you believe technologies such as IDS are still relevant and useful as
there is a push toward SaaS and cloud-based solutions.
(b) Discuss from your perspective how cloud-based services change incident response, for better
or worse, and determine what you believe to be the greatest preparedness concern with cloud-
based services.
PLEASE NO COPY AND PASTING
Solution
(a)
IDS - Intrusion Detection systems
To prevent threat over once computer security firewall and encryption techniques were invented.
Although, the intruders could still hack one\'s system. To prevent this, Intrusion detection
systems were introduced which provide the following advantages:
1.Continuously monitors and analyzes the activities of the user and system.
2.Uses pattern recognition for recognizing typical kinds of attacks.
3.Tracks user policy violations.
3. Recognizes abnormal activities through pattern recognition.
SaaS: Software as a Service
1.SaaS is a model for Software licensing and delivery which is hosted on internet on subscription
basis and made available for the internet users.
2. It is one of the categories of cloud computing, where in instead of simply installing and
maintaining software you keep it over the internet which is easily maintainable from complex
software management.
Intrusion detection systems are still relevant to protect one\'s private data although Cloud
computing exists.
(b) Security and Incident response is a major subject which many vendors try to overcome.
At high level, the process defines six stages of incident response :
Preparation,identification,containment,eradication,recovery,lessons learned.
Five challenges or risks or concerns faced with cloud computing:
1.Privacy and Security.
2.Cloud transition.
3.Data access
4.Service quality
5.Data accessibility..
Gandalf the Grey started in the Forest of Mirkwood at a point with co.pdf
Gandalf the Grey started in the Forest of Mirkwood at a point with coordinates (-3, -1) and
arrived in the Iron Hills at the point with coordinates (-1,3). If he began walking in the direction
of the vector v = 5i + lj and changes direction only once, when he turns at a right angle, what are
the coordinates of the point where he makes the turn.
Solution
Geometrically, the vector 5i + 1j corresponds to a slope of 1/5.
So, at the right angle where Gandalf changes direction, the slope becomes -5.
Let\'s write out the equations.
Point (-3, -1), slope 1/5 => y + 1 = (1/5)(x + 3) or y = (1/5)(x + 3) - 1 or y = (1/5)(x - 2)
Point (-1, 3), slope - 5 => y - 3 = (-5)(x + 1) or y = - 5x - 2.
Now, we find the intersecting point of these two curves
(1/5)(x - 2) = -5x - 2
==> x = - 4/13
Substituting this into either equation yields y = - 6/13.
So, the desired point is ( - 4/13, - 6/13)..
Explain exploratory factor analysis and give real life examplesS.pdf
Explain exploratory factor analysis and give real life examples
Solution
In multivariate statistics, exploratory factor analysis (EFA) is a statistical method used to
uncover the underlying structure of a relatively large set of variables. EFA is a technique within
factor analysis whose overarching goal is to identify the underlying relationships between
measured variables.[1] It is commonly used by researchers when developing a scale (a scale is a
collection of questions used to measure a particular research topic) and serves to identify a set of
latent constructs underlying a battery of measured variables.[2] It should be used when the
researcher has no a priori hypothesis about factors or patterns of measured variables.[3]Measured
variables are any one of several attributes of people that may be observed and measured. An
example of a measured variable would be the physical height of a human being. Researchers
must carefully consider the number of measured variables to include in the analysis.[2] EFA
procedures are more accurate when each factor is represented by multiple measured variables in
the analysis.
Following is the set of EFA examples,
. • 4.1: Exploratory factor analysis with continuous factor indicators
• 4.2: Exploratory factor analysis with categorical factor indicators
• 4.3: Exploratory factor analysis with continuous, censored, categorical, and count factor
indicators*
• 4.4: Exploratory factor mixture analysis with continuous latent class indicators
• 4.5: Two-level exploratory factor analysis with continuous factor indicators
• 4.6: Two-level exploratory factor analysis with both individual- and cluster-level factor
indicators
• 4.7: Bi-factor exploratory factor analysis with continuous factor indicators
* Example uses numerical integration in the estimation of the model. This can be
computationally demanding depending on the size of the problem..
Drag the labels to the correct category to demonstrate your understan.pdf
Drag the labels to the correct category to demonstrate your understanding of cell walls in
different types of microbes. Protozoans Helminths Bacteria Fungi
Solution
Protozoans- These are classified under kingdom Protist.
Helminths-These are classified under kingdom animalia.
Fungi-These are classified under kingdom fungi.
Bacteria- Prokaryotes..
Do you agree that the Berg Letter, research moratorium, and Asilomar.pdf
Do you agree that the Berg Letter, research moratorium, and Asilomar Conference were
necessary before further recombinant DNA research was conducted? Why or why not?
Solution
Paul Berg, an American biochemist and professor emeritus at Stanford University witnessed
firsthand the history of recombinant DNA research and regulation, having been in the forefront
of both movements since he was a young man. He gained early recognition and influence when
he delineated the key steps in which DNA produces proteins.
In the mid-1970s, the National Academy of Sciences asked Berg to explore the safety of
recombinant DNA technology. He responded with the historic \"Berg letter,\" calling for a
moratorium on recombinant DNA research until safety issues could be addressed. He was one of
the key organizers of the international forum on recombinant DNA technology, the Asilomar
Conference. The ensuing dialogue resulted in the National Institutes of Health guidelines
published a year later, a milestone of responsible self-regulation in science.
The Asilomar Conference on Recombinant DNA was an influential conference organized by
Paul Berg to discuss the potential biohazards and regulation of biotechnology, held in February
1975 at a conference center at Asilomar State Beach.A group of about 140 professionals
(primarily biologists, but also including lawyers and physicians) participated in the conference to
draw up voluntary guidelines to ensure the safety of recombinant DNA technology. The
conference also placed scientific research more into the public domain, and can be seen as
applying a version of the precautionary principle.
The effects of these guidelines are still being felt through the biotechnology industry and the
participation of the general public in scientific discourse. Due to potential safety hazards,
scientists worldwide had halted experiments using recombinant DNA technology, which entailed
combining DNAs from different organisms. After the establishment of the guidelines during the
conference, scientists continued with their research, which increased fundamental knowledge
about biology and the public’s interest in biomedical research.
During the conference, the principles guiding the recommendations for how to conduct
experiments using this technology safely were established. The first principle for dealing with
potential risks was that containment should be made an essential consideration in the
experimental design. A second principle was that the effectiveness of the containment should
match the estimated risk as closely as possible.
The conference also suggested the use of biological barriers to limit the spread of recombinant
DNA. Such biological barriers included fastidious bacterial hosts that were unable to survive in
natural environments. Other barriers were nontransmissible and equally fastidious vectors
(plasmids, bacteriophages, or other viruses) that were able to grow in only specified hosts.
In addition to biological barriers, the co.
Brief Concepts and Questions EXAM 2 Chapter 8 DNA RNA Protein What i.pdf
Brief Concepts and Questions EXAM 2 Chapter 8: DNA RNA Protein What is DNA? a
phosphate Structure of DNA: Building blocks are called nucleotides Each nucleotide is
composed of three br uithofenas bee. What makes DNA so special? Provide 4 reasons, below
DNA DNA (Replication): Where does DNA replication take place? When does DNA replication
take place? Explain steps involved in DNA replication: DNA RNA Protein (Gene Expression)
Involves 2 processes: 1. Transcription 2. Translation Explain the Synthesis of Proteins (Gene
Expression): o DNA RNA Protein What is RNA? What is \"codon What is \"anticodon\" What is
a protein molecule? DNA mutation; Change in nucleotide bases of DNA Duplex Point mutation
Frame shift mutation
Solution
Question
Answer
Where does DNA replication take place:
It takes place in the nucleus in case of eukaryotic cells and in the cytoplasm in case of
prokaryotic cells
When does DNA replication take place:
DNA replication occurs during the S-phase during cell cycle, so that cell can make an extra copy
of genetic material.
Explain steps involved in DNA replication:
Initiation: During initiation, the proteins will bind to the origin of replication; helicase unwinds
the DNA helix which results in the formation of two replication forks.
Elongation: A RNA primer sequence will be added to this the DNA pol III will add the
nucleotides in 5’ to 3’ direction and chain will elongate.
Termination: In case of bacteria, termination of replication occurs whenever two replication
forks meet each other from the opposite end of the parental chromosome.
Transcription
Gene expression first step is transcription, here a particular segment of DNA will be copied into
RNA with the help of the enzyme RNA polymerase
Translation
Translation is the final step of the gene expression. Here mRNA will be used to synthesize the
polypeptide chain. The information present in the mRNA in the form of codon will code for the
amino acids needed for polypeptide chain synthesis.
What is RNA?
RNA is ribonucleic acid and is found in all living cells. It acts as the messenger carrying
instructions from DNA for the synthesis of proteins.
Few viruses will have RNA as their genetic material.
What is codon?
Codon is a sequence of three nucleotides and they together form a unit of genetic code in either
DNA or RNA.
What is anticodon?
It is found on tRNA and it is a sequence of three nucleotides which forms a genetic code on
tRNA, and these anticodon is complementary to the codons found on messenger RNA.
What is a protein molecule?
During translation, when amino acids are added in a sequential manner, the condensation of
amino acids will form a peptide bond in between them and finally forms a polypeptide chain. It
is the DNA through mRNA directs the protein synthesis.
Point mutation
In point mutation, only one or very few nucleotides will be affected or mutated in a gene
sequence.
Frame shift mutation
Either insertions or deletion can result in frame shift mutation, due to th.
A population has two alleles for a particular gene, A and a, and is .pdf
A population has two alleles for a particular gene, A and a, and is in Hardy-Weinberg
equilibrium. The frequency of allele a is 0.2. What is the frequency of individuals with AA
genotype?
0.2
0.4
0.64
0.04
0.8
Solution
C). 0.64
Hardy and Weinberg mathematically proved that in a population, all dominant and recessive
alleles comprise all alleles for that gene.
This was mathematically represented as p+ q = 1.0
Where,
p = frequency of dominant alleles
q = frequency of recessive alleles.
Given that, the frequency of a = 0.2
As the population is in H-W equilibrium, the frequency of A = 1-0.2 = 0.8
The frequency of AA genotype = 0.8*0.8 = 0.64.
A 12-year-old male complained of strange odors before loss of consci.pdf
A 12-year-old male complained of strange odors before loss of consciousness and a major tonic-
clonic seizure. During the seizure, his parents rushed him to the emergency room of a local
hospital. On arrival at the hospital, he appeared to be asleep. Studies at the hospital showed
routine laboratory work within normal limits (WNL), lumbar puncture (CSF) was WNL, no
evidence of skull fracture on x-ray study was revealed, and an electroencephalograph (EEG)
showed no abnor malities. How would you interpret the episode and findings?
Solution
Charecterized by sudden loss of consciousness ,chaotic muscle movement and tone, and apnea
Tonic phase:Bilateral muscle rigidity
Clonic phase: muslce contraction and relaxation lasting 1 to 3 minutes
Signs:
Tachycardia, hyerpventilation,sweating, and intense salivation
Last 3-5 minutes
postictal state(5 to 20 mintues)follows
-gradual return to consciousness.
2.Which of the following statements about evolution is falseNonrand.pdf
2.Which of the following statements about evolution is false?Nonrandom mating can change the
genetic makeup of populations over time.Genetic drift can change the genetic makeup of
populations over time.Natural selection is one of several mechanisms of evolution.Populations,
not individuals, evolve.Evolution is goal-directed.2.Which of the following statements about
evolution is false?Nonrandom mating can change the genetic makeup of populations over
time.Genetic drift can change the genetic makeup of populations over time.Natural selection is
one of several mechanisms of evolution.Populations, not individuals, evolve.Evolution is goal-
directed.
Solution
Note:.
can someone explain in a easy way chromosome walking I know its w.pdf
can someone explain in a easy way chromosome walking? I know it\'s with overlapping
fragments and a gen nearby the gen you want to find, but I don\'t clearly understand it, I have
images in my book, but an image with steps and explanation would be nice,
Solution
Chromosome Walking:
Chromosome walking is positional cloning of gene sequences or upstream promoter sequences.
Eukaryotic DNA contains a lot of junk DNA between the genes. The technique relies upon
screening of gDNA library for overlapping fragments in a sequential manner to tap the upstream
or downstream region.
Procedure:
1. Isolation of Genomic DNA: First of all isolate genomic DNA (gDNA) of the test organism
and then partially digest it with some restriction enzyme, say Hind III that would yield unequal
random fragments.
2. Cloning of the Digested into E.coli: Clone the digested gDNA into a cloning vector (between
the Hind III sites in this case) to prepare a gDNA library of the organism. Each colony of the
library would have only one clone.
3. Preparation of marker probe: Prepare a radioactively labelled probe from the neighboring gene
(say A), from where you want to start walking.
4. Colony Hybridization: Hybridize the probe with your gDNA library. The probe would show
the signal in the colonies carrying DNA in and around your gene.
5. Sequencing: Pick the colonies showing the signal and isolate the plasmid DNA and sequence
each using the plasmid-specific primers. You would get only one sequence per bacterial colony
of your library.
The sequence, thus obtained must be analyzed. Select longer sequences towards the direction of
your walk.
6. Preparation of the probe: Prepare new probes from the region towards your target gene.
7. Repeat the steps 4-6 till you get the clone and sequence of your target gene..
9. When a bacteriophage inserts its DNA into a bacterial chromosome,.pdf
9. When a bacteriophage inserts its DNA into a bacterial chromosome, the inserted
bacteriophage DNA is called _____
(a) prophage
(b) lytogenic
(c) virulent
Solution
a.Prophage.
The genetic material of a bacteriophage ,incorporated in to the genome of bacterium or host ,now
it is ready to prouce phages if specifically activated.
This stage of Bacteriophage life cycle called as Prophage..
Write an informal negation for each of the following statements. Be .pdf
Write an informal negation for each of the following statements. Be careful to avoid negations
that are ambiguous. a. All dogs are friendly. b. All people are happy. c. Some suspicions were
substantiated. d. Some estimates are accurate.
Solution
(a)
these is a dog which is not friendly
(b)
Some people are not happy
(c)
these was a ssuspicions which was not substantiated
(d)
these is some estimates which are wrong.
Write a C++ program which generates and displays a random walk acros.pdf
Write a C++ program which generates and displays a random walk across a grid. A 2-
dimensional array of characters will be used for this grid. The cells visited by the walk will be
labeled by capital letters with A being the first cell, B the cell moved to from A, and so on,
ending either when Z is reached or when it is not possible to move to another cell.
The following libraries, constant definition, and type definition are needed in the program:
#include
#include // for srand and rand
#include // for time
using namespace std;
const int SIZE = 10;
typedef char Grid[SIZE][SIZE]
The grid must be set to have all its elements be the character ’.’ before the walk is begun.
The program must contain the following function prototypes, the main function, and the function
definitions for all the other functions:
// Function prototypes
bool can_move_up(int i, int j, Grid walk);
bool can_move_down(int i, int j, Grid walk);
bool can_move_left(int i, int j, Grid walk);
bool can_move_right(int i, int j, Grid walk);
void init_array(Grid walk);
void generate_random_walk(Grid walk);
void print_array(Grid walk);
int main(void)
{
Grid walk; // the grid in which the random walk occurs
srand((unsigned) time(NULL));
init_array(walk);
generate_random_walk(walk);
print_array(walk);
return 0;
}
The four functions can_move _* must return whether or not it is possible to move in the
particular direction in the grid. For example, a function call like if (can_move_up(i, j, walk))
could be used to determine whether or not it is possible to move up. The rule for a move being
possible is that if it doesn’t go outside of the grid and isn’t already part of the walk, then that
move is possible. The function init_array must set all characters in the grid to ’.’
The function generate_random_walk must modify the cells of grid by changing ones that are
part of the walk to letters.
The function print_array must print the resulting grid in the form shown in the sample run
below.
In the function generate_random_walk, at each step of the walk, randomly select one of the 4
directions and try to move in that direction. If that move isn’t available, try the other moves in
turn until one is possible. If all 4 directions are blocked, the walk must terminate, and the
program must continue and print out the walk. To select one of the 4 directions randomly, use
the expression: rand() % 4
The function rand returns an integer, and finding the remainder of this integer divided by 4
results in getting one of the numbers 0, 1, 2, or 3. Use these 4 numbers to represent the 4
directions up, down, left and right.
The call to the function srand from the main function seeds the pseudo-random number
generator used by rand so that different sequences of numbers will be generated by calls to rand
each time the program is run.
Solution
#include //input output functions in C++
#include //rand() and srand() functions
#include //time() function for creating seed
#include
#include // for srand and rand
#inclu.
true or false1- The grouped frequency table should ideally have.pdf
true or false???
1- The grouped frequency table should ideally have 10 class intervals
2- The width of the intervals should be half of the number of scores
3- Each score can be represented in one and only one interval
4- The intervals should end with the highest number in the sample
Solution
1.True
2.False
3.True
4.False.
Answer the questions that follow the VHDL code given below. LIBRARY .pdf
Answer the questions that follow the VHDL code given below. LIBRARY ieee; USE
ieee.std_logic_1164.ALL; ENTITY ex1_IS PORT (n, cp, n rd: IN st_logic; q: BUFFER integer
RANGE 0 TO 15); END ex1; ARCHITECTURE are OF ex1 IS BEGIN PROCESS (n_cp,
n_rd) BEGIN IF (n_rd = \'0\') THEN q
Solution
a)n_rd- used to reset the counter
n_cp-used as clock to the counter
b)4 binary bits are there in q;
Q stores 0 t0 15 number so
log2(16)=4.
The coding region of a BACTERIAL gene has the following sequence A.pdf
The coding region of a BACTERIAL gene has the following sequence:
\"AGAGCCAACATGTCGAGCATATGCTGATATACG\' Draw and label the DNA (including
ALLEABIS of the gene, sequence information) The info you need is in the sides. Label the
template/coding strands Label 5\' and 3\' ends of each DNA strand Label where RNA
polymerase binds \"Transcribe\" this gene into mRNA, as shown in Bb video. Write the mRNA
nucleotide sequence, starting at the correct nucleotide, starting at the correct nucleotide Show
the 5\' and 3\' ends of the RNA. Use the codon chart to \"translate\" the mRNA into a protein,
and write the correct amino acid sequence of the protein. Only translate the part of the gene that
a cell would translate. Label the N-terminus and C-terminus of the protein. If the sequence of
the gene were changed to: AGACCCAACATGTCGGACATATGCTGATATACG what would
happen to the properties of the protein? Explain why write out the entire process of bacterial
transcription step by step in the correct order, in detail.
Solution
4. Process of Bacterial Transcription:
Bacterial transcription is the process in which mRNA transcripts of genetic material in bacteria
are produced; it is further translated to produce proteins. Unlike in eukaryotes, bacterial
transcription and translation can occur simultaneously in the cytoplasm.
RNA polymerases are enzymes that transcribe DNA into RNA. Using a DNA template, RNA
polymerase builds a new RNA molecule through base pairing. For instance, if there is a G in the
DNA template, RNA polymerase will add a C to the new, growing RNA strand.
Three steps are involved in bacterial transcription:
Initiation: To begin transcribing a gene, RNA polymerase binds to the DNA of the gene at a
region called the promoter. Basically, the promoter tells the polymerase where to locate on the
DNA and begin transcribing.
Elongation:
Once RNA polymerase is in position at the promoter, the next step of
transcription—elongation—can begin. Basically, elongation is the stage when the RNA strand
gets longer, thanks to the addition of new nucleotides.
During elongation, RNA polymerase \"walks\" along one strand of DNA, known as the template
strand, in the 3\' to 5\' direction. For each nucleotide in the template, RNA polymerase adds a
matching (complementary) RNA nucleotide to the 3\' end of the RNA strand.
Termination:
RNA polymerase will keep transcribing until it gets signals to stop. The process of ending
transcription is called termination, and it happens once the polymerase transcribes a sequence of
DNA known as a terminator.
There are two major termination strategies found in bacteria: Rho-dependent and Rho-
independent.
In Rho-dependent termination, the RNA contains a binding site for a protein called Rho factor.
Rho factor binds to this sequence and starts \"climbing\" up the transcript towards RNA
polymerase.
When it catches up with the polymerase at the transcription bubble, Rho pulls the RNA transcript
and the template DNA strand apart, releasing .
Regarding the ABO blood group system in humans, if an individual is g.pdf
Regarding the ABO blood group system in humans, if an individual is genetically I^B i and yet
expresses the O blood type, which genotype is he or she likely to have? Hh HH hh i^A i i^B i
none of the above are correct.
Solution
Option C is the correct answer. This hh is also known as Bombay blood group. These individuals
do not express H antigen. Thus these individuals are unable to form either of A or B antigen on
their RBCs. Usual test of these individuals for blood group shows them as having O blood group..
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- 1. The weak D phenotype is thought to arise from several mechanisms. Which one of the following statements is false? C gene is in trans position to D gene, this results in a weakening influence on the D gene, examples CDe/Cde and cDe/Cde. Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in caucasians. Also occurs in the Ro genotype in certain racial groups more commonly. The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these pieces is missing. The red cells are said to have a partial D. D deletion genes gives rise to weakened expression of D antigen. Rh null cells lack: MNSs antigens. Lewis antigens. Normal oxygen-carrying capacity. Rh antigens. There are two closely related and closely linked genes. Enzyme techniques, in conjunction with an IgM anti-D. Increasing the time for the incubation of the saline phase. Performance of an IAT using an IgG anti-D. Don't need to do it in a blood bank. C gene is in trans position to D gene, this results in a weakening influence on the D gene, examples CDe/Cde and cDe/Cde. Weak D antigen is genetically transmitted. This can occur rarely in R1 and R2 genotypes in caucasians. Also occurs in the Ro genotype in certain racial groups more commonly. The D antigen consists of multiple pieces (is a mosaic). Occasionally one or more of these pieces is missing. The red cells are said to have a partial D. D deletion genes gives rise to weakened expression of D antigen. Solution The weak D phenotype is thought to arise from several mechanisms. Which one of the following statements is false? D deletion genes gives rise to weakened expression of D antigen.
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