8. Define o jitney o tanker ship o STO o RORO o IFR o inland waterwa.pdf
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8. Define: o jitney o tanker ship o STO o RORO o IFR o inland waterway o range lights Solution Jitney is a vehicle or a bus which carries passengers for low rate. tanker ship: Which carries liquids and gases in bulk amount. STOL: It is a aricraft which require only short runway to takeoff or land RORO: Roll on roll off ships are used to carry the wheeled vehicles. Inland waterways: transportation through riverwater. range lights: There used to navigate..
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With which portion of an epithelial cell in the stomach would food be.pdf
With which portion of an epithelial cell in the stomach would food be in contact? basal surface
basement membrane tight junctions apical surface gap junctions Which of the following
relationships between a tissue and its general function is not correct? Epithelial: control Nerve:
control Connective: support Muscle: movement Goblet cells are found in which of these
epithelia. Stratified squmous Simple squmaous Transitional Simple columnar
Solution
1. apical surface
The stomach surface is lined with mucus secreting columnar epithelial cells. A single
cell structure consists of three surfaces-apical, lateral and basal.
The apical surface faces the intestinal lumen. Extending from the apical surface are
numerous fingerlike projections (100 nm in diameter) called microvilli, referred as the brush
border due to their appearance. Microvilli greatly increase the area of the apical surface,
enhancing the absorptive and transportive capacity of the intestinal epithelium. Overlying the
brush border, are the oligosaccharide side chains of integral membrane glycoproteins,
glycolipids, and enzymes (glycocalyx) catalyzing the final digestion of ingested carbohydrates
and proteins. The resultant monosaccharides and amino acids produced are transported across the
intestinal epithelium and eventually into the bloodstream.
The lateral surface encloses tight junctions, adherens junctions, spotdesmosome, gap
junction, intermediate filament, and hemidesmosome.
-Tight junctions, prevent diffusion of substances between the intestinal lumen and the blood via
the extracellular space between cells. They function as a selective and semipermeable barrier
between apical and basolateral compartments of the epithelial layer. They facilitate the passage
of small ions and water-soluble solutes through the paracellular space while preventing the
passage of luminal antigens, microorganisms and their toxins.
-Gap junctions are door-like connections between each epithelial cells.They allow movement of
small molecules and ions between the cytosol of adjacent cells.
-The remaining three types of junctions, adherens junctions, spot desmosomes, and
hemidesmosomes are critical to cell-cell and cell-matrix adhesion.
The basal surface of the epithelial cells is the basal lamina, consisting of a fibrous
network of collagen and proteoglycans that supports the epithelial cell layer.
2. epithelial:control
Epithelial cells line the body cavities and surface of organs and blood vessels. Their functions
include secretion, selective absorption, protection, transcellular transport, and sensing.
3. simple columnar
Goblet cells are found scattered among the epithelial cells, mostly in the respiratory and
intestinal tracts. These cells are simple columnar epithelial cells with a narrow base and
expanded apical portion (glandular shape). They secrete mucus which protect against stress and
chemical damage, and destroy the microorganisms and particulate matter..
Write the interface (.h file) of a class Accumulator containing A d.pdf
Write the interface (.h file) of a class Accumulator containing: A data member named sum of
type integer. A constructor accepting no parameters. A function named getSum that accepts no
parameters and returns an integer. A function named add that accepts an integer parameter and
returns no value.
Solution
Accumulator.h
class Accumulator{
private:
int sum;
public:
Accumulator();
int getSum();
void add(int x);
};.
You are carrying out PCRs in lab. Why do you not need helicase, SSB,.pdf
You are carrying out PCRs in lab. Why do you not need helicase, SSB, primase, ligase, sliding
clamp?
Solution
1. Helicase: Helicase is an eznyme that plays a crucial role in seperating the DNA double helix
by unwinding it.
2. SSB: It stands for Single-Stranded DNA Binding Protein (SSB). It binds with high affinity to
single-stranded DNA and destabilizes helical duplexes. This helps DNA polymerases to bind to
substrates easily. SSB do not bind to double stranded DNAs.
3. Primase: Primase is an enzyme that synthesize primers (short stretches of DNA for the DNA
synthesis to proceed) in vivo. In In a PCR experiment, two specific primers are designed and
added in the reaction for amplifying specific sequences.
4. Ligase: This enzyme catalyze the DNA repair process by joining of ends of DNA molecules.
In these reactions, this enzyme links the phosphate to a neighboring sugar molecule, thus
forming the backbone of DNA.
5. Sliding clamp: This clamp protein binds to DNA polymerase (enyme synthesizing the DNA
chain), and prevents it from dissociation from the template. Thus it acts as a promoting factor in
DNA replication process. The clamp-polymerase interactions are stronger and more specific. It
increases the rate of reaction progression..
Write your own definition of cloning. No plagarism!SolutionClo.pdf
Write your own definition of cloning. No plagarism!
Solution
Cloning:-
Exact copying or Repilcating any thing is called as cloning.
Per supose in Biology Cloning is defind as --- A group of cells or organisms that are descended
from and genetically identical to a single progenetor, such as a bacterial colony whose members
arose from a single cell.
Write appropriate SQL DDL statements (Create Table Statements) for d.pdf
The document provides SQL DDL statements to define the schema for a relational database about researchers, experiments, microarrays, and the relationship between experiments and microarrays. It includes CREATE TABLE statements that define primary keys and foreign keys to link the tables for researcher name and ID, experiment ID and date, microarray ID and attributes, and a join table to contain the many-to-many relationship between experiments and microarrays.
Write a class called Student that extends the provided Person class..pdf
Write a class called Student that extends the provided Person class. Ensure that your student
class is a concrete class. The foo method for a person should display (to standard output) the
person\'s age and name in a nice way (e.g, \"Nigel\'s age is 14\"). When a person makes a noise it
is their name repeated N times where N is the person\'s age. public abstract class Person
implements Noisy Person Class public String name; 3 ublic int age; s public abstract void foo);
public interface Noisy 2 String noise) 1 public interface Noisy{ Noisy Interface 1 public
interface Comparable 2 i Comparable Interface int compareTo(Type other); 1 public class Date 2
int day, month, year; Date Class public Date(int d, int m, int y){ dayd; month - m; year-y;
Solution
public class Student extends Person{
Student(String name, int age){
this.name = name; // initialising the student properties
this.age = age;
}
public void foo(){
if(isNoisy()){ //check if person makes noise
for(int i=0 ; i< age; i++ ){
System.out.print(\" \"+name+\" \");
}
}else{
System.out.println(name+\"\\\'s age is \"+age);
}
}
}
// Note: Information about the Noisy class is required..
why NADH and FADH2 are a type of energy currencySolutionEnergy.pdf
why NADH and FADH2 are a type of energy currency
Solution
Energy used by the body for its many activities ultimately comes from the chemical energy in
our food.
The chemical energy in our food is converted to reducing agents NADH and FADH2 .
These reducing agents are then used to make ATP.
ATP stores chemical energy ,so that it is available to the body in a readily accessible form.
The energy needed by the body is coming from NADH and FADH2, that why they are type of
energy currency..
Who is D.L ShellSolutionHi friend,Donald L. Shell (March 1,.pdf
Who is D.L Shell?
Solution
Hi friend,
Donald L. Shell (March 1, 1924 – November 2, 2015) was an American computer scientist who
designed the Shell sort sorting algorithm. He acquired his Ph.D. in Mathematics from the
University of Cincinnati in 1959, and published the shell sort algorithm in the Communications
of the ACM in July that same year.
YOu can read on: https://en.wikipedia.org/wiki/Donald_Shell.
Recommended
With which portion of an epithelial cell in the stomach would food be.pdf
With which portion of an epithelial cell in the stomach would food be in contact? basal surface
basement membrane tight junctions apical surface gap junctions Which of the following
relationships between a tissue and its general function is not correct? Epithelial: control Nerve:
control Connective: support Muscle: movement Goblet cells are found in which of these
epithelia. Stratified squmous Simple squmaous Transitional Simple columnar
Solution
1. apical surface
The stomach surface is lined with mucus secreting columnar epithelial cells. A single
cell structure consists of three surfaces-apical, lateral and basal.
The apical surface faces the intestinal lumen. Extending from the apical surface are
numerous fingerlike projections (100 nm in diameter) called microvilli, referred as the brush
border due to their appearance. Microvilli greatly increase the area of the apical surface,
enhancing the absorptive and transportive capacity of the intestinal epithelium. Overlying the
brush border, are the oligosaccharide side chains of integral membrane glycoproteins,
glycolipids, and enzymes (glycocalyx) catalyzing the final digestion of ingested carbohydrates
and proteins. The resultant monosaccharides and amino acids produced are transported across the
intestinal epithelium and eventually into the bloodstream.
The lateral surface encloses tight junctions, adherens junctions, spotdesmosome, gap
junction, intermediate filament, and hemidesmosome.
-Tight junctions, prevent diffusion of substances between the intestinal lumen and the blood via
the extracellular space between cells. They function as a selective and semipermeable barrier
between apical and basolateral compartments of the epithelial layer. They facilitate the passage
of small ions and water-soluble solutes through the paracellular space while preventing the
passage of luminal antigens, microorganisms and their toxins.
-Gap junctions are door-like connections between each epithelial cells.They allow movement of
small molecules and ions between the cytosol of adjacent cells.
-The remaining three types of junctions, adherens junctions, spot desmosomes, and
hemidesmosomes are critical to cell-cell and cell-matrix adhesion.
The basal surface of the epithelial cells is the basal lamina, consisting of a fibrous
network of collagen and proteoglycans that supports the epithelial cell layer.
2. epithelial:control
Epithelial cells line the body cavities and surface of organs and blood vessels. Their functions
include secretion, selective absorption, protection, transcellular transport, and sensing.
3. simple columnar
Goblet cells are found scattered among the epithelial cells, mostly in the respiratory and
intestinal tracts. These cells are simple columnar epithelial cells with a narrow base and
expanded apical portion (glandular shape). They secrete mucus which protect against stress and
chemical damage, and destroy the microorganisms and particulate matter..
Write the interface (.h file) of a class Accumulator containing A d.pdf
Write the interface (.h file) of a class Accumulator containing: A data member named sum of
type integer. A constructor accepting no parameters. A function named getSum that accepts no
parameters and returns an integer. A function named add that accepts an integer parameter and
returns no value.
Solution
Accumulator.h
class Accumulator{
private:
int sum;
public:
Accumulator();
int getSum();
void add(int x);
};.
You are carrying out PCRs in lab. Why do you not need helicase, SSB,.pdf
You are carrying out PCRs in lab. Why do you not need helicase, SSB, primase, ligase, sliding
clamp?
Solution
1. Helicase: Helicase is an eznyme that plays a crucial role in seperating the DNA double helix
by unwinding it.
2. SSB: It stands for Single-Stranded DNA Binding Protein (SSB). It binds with high affinity to
single-stranded DNA and destabilizes helical duplexes. This helps DNA polymerases to bind to
substrates easily. SSB do not bind to double stranded DNAs.
3. Primase: Primase is an enzyme that synthesize primers (short stretches of DNA for the DNA
synthesis to proceed) in vivo. In In a PCR experiment, two specific primers are designed and
added in the reaction for amplifying specific sequences.
4. Ligase: This enzyme catalyze the DNA repair process by joining of ends of DNA molecules.
In these reactions, this enzyme links the phosphate to a neighboring sugar molecule, thus
forming the backbone of DNA.
5. Sliding clamp: This clamp protein binds to DNA polymerase (enyme synthesizing the DNA
chain), and prevents it from dissociation from the template. Thus it acts as a promoting factor in
DNA replication process. The clamp-polymerase interactions are stronger and more specific. It
increases the rate of reaction progression..
Write your own definition of cloning. No plagarism!SolutionClo.pdf
Write your own definition of cloning. No plagarism!
Solution
Cloning:-
Exact copying or Repilcating any thing is called as cloning.
Per supose in Biology Cloning is defind as --- A group of cells or organisms that are descended
from and genetically identical to a single progenetor, such as a bacterial colony whose members
arose from a single cell.
Write appropriate SQL DDL statements (Create Table Statements) for d.pdf
The document provides SQL DDL statements to define the schema for a relational database about researchers, experiments, microarrays, and the relationship between experiments and microarrays. It includes CREATE TABLE statements that define primary keys and foreign keys to link the tables for researcher name and ID, experiment ID and date, microarray ID and attributes, and a join table to contain the many-to-many relationship between experiments and microarrays.
Write a class called Student that extends the provided Person class..pdf
Write a class called Student that extends the provided Person class. Ensure that your student
class is a concrete class. The foo method for a person should display (to standard output) the
person\'s age and name in a nice way (e.g, \"Nigel\'s age is 14\"). When a person makes a noise it
is their name repeated N times where N is the person\'s age. public abstract class Person
implements Noisy Person Class public String name; 3 ublic int age; s public abstract void foo);
public interface Noisy 2 String noise) 1 public interface Noisy{ Noisy Interface 1 public
interface Comparable 2 i Comparable Interface int compareTo(Type other); 1 public class Date 2
int day, month, year; Date Class public Date(int d, int m, int y){ dayd; month - m; year-y;
Solution
public class Student extends Person{
Student(String name, int age){
this.name = name; // initialising the student properties
this.age = age;
}
public void foo(){
if(isNoisy()){ //check if person makes noise
for(int i=0 ; i< age; i++ ){
System.out.print(\" \"+name+\" \");
}
}else{
System.out.println(name+\"\\\'s age is \"+age);
}
}
}
// Note: Information about the Noisy class is required..
why NADH and FADH2 are a type of energy currencySolutionEnergy.pdf
why NADH and FADH2 are a type of energy currency
Solution
Energy used by the body for its many activities ultimately comes from the chemical energy in
our food.
The chemical energy in our food is converted to reducing agents NADH and FADH2 .
These reducing agents are then used to make ATP.
ATP stores chemical energy ,so that it is available to the body in a readily accessible form.
The energy needed by the body is coming from NADH and FADH2, that why they are type of
energy currency..
Who is D.L ShellSolutionHi friend,Donald L. Shell (March 1,.pdf
Who is D.L Shell?
Solution
Hi friend,
Donald L. Shell (March 1, 1924 – November 2, 2015) was an American computer scientist who
designed the Shell sort sorting algorithm. He acquired his Ph.D. in Mathematics from the
University of Cincinnati in 1959, and published the shell sort algorithm in the Communications
of the ACM in July that same year.
YOu can read on: https://en.wikipedia.org/wiki/Donald_Shell.
Which of the following correctly states the functions of the roug.pdf
Which of the following correctly states the functions of the rough endoplasmic reticulum ? lipid
metabolism, cholesterol synthesis, and synthesis of the lipid components of lipoproteins
absorption, synthesis, and transport of fats. storage of calcium ions synthesis of the cell\'s
membranes
Solution
As per the given options rough endoplasmic reticulum functions are :-
2) Absorbtion, synthesis and transport of fats & 4) Synthesis of cell membranes
Generally, the eucaryotic cell contains two types of endoplasmic reticulum
i) Rough Endoplasmic Reticulum.
Rough Endoplasmic Reticulum why named as rough means it contains ribosomes on it so, the
surface of the reticulum looks rough. The main function of the rough ER is protein synthesis due
to the ribosomes.
Rough ER through the membrane contact sites contact with the other organells allowing the
transfer of lipids and other small molecules.
The intestinal cells, pancreatic cells, which involves in the absorption process contains more
rough ER why because the proteins synthesized by the Rough ER involved in the intracellular
transport of resynthesized lipids.So the rough ER indirectly helps in the absorption process.
ii)Smooth Endoplasmic Reticulum:-
Due to the absence of ribosomes it looks smooth so named as smooth ER. The main role is lipid
metabolism, and synthesis of carbohydrates and steroid harmones and storage of calcium ions
So, in the given question the first option(Lipid metabolism, cholesterol synthesis, synthesis of
lipid components of lipo proteins) indicates smooth Endoplasmic Reticulum functions
2nd option Absorbtion, synthesis and transport of fats indicates Rough Endoplasmic Reticulum
functions
3rd Option Indicates(Storage of Calcium ions) smooth Endoplasmic Reticulum functions
4th Option is synthesis of cell membrane, For this function the main role of rough Endoplasmic
reticulum is during the lipid transport the replenishment of new membranes is done by the
morphological change of rough endoplasmic reticulum involved..
What is isomerism How many kinds are there Define each kind and.pdf
What is isomerism? How many kinds are there? Define each kind and explain with examples?
What is isomerism? How many kinds are there? Define each kind and explain with examples?
Solution
The process of isomerism involves the isomers .These are basically defined as the molecules
which have the same molecular formula but they usually possess a different arrangement of the
atoms in space. They are of two types:
1.Constitutional isomers :They are defined as the compounds in which the various atoms are
connected or bonded in a different manner .They are also called as structural isomers
2.Stereoisomers:On the other hand ,they are the compounds which differ on the basis of
orientation in space.They are of two types
a)Enantiomers :They are the compounds which are usually a mirror image of one another and
thehese mirror images are also non-superimposable
b)Diastereomers :They are isomers which are not mirror images of one another.Moreover these
compounds are non-superimposable. They are of two types: cis and trans
->Cis-isomers are the isomers which usually contain atoms which are oriented on the same side
of a bond.
->On the other hand the Trans-isomers are the compounds that have atoms oriented on opposite
sides of a bond.
What is the definition of the words, and state the diffe.pdf
What is the definition of the words, and state the difference between them if there is:
1- D-Flip Flop
2- Register
3- Equality
4- Counter
5- Adder
6- Multiplexer
What is the definition of the words, and state the difference between them if there is:
1- D-Flip Flop
2- Register
3- Equality
4- Counter
5- Adder
6- Multiplexer
What is the definition of the words, and state the difference between them if there is:
1- D-Flip Flop
2- Register
3- Equality
4- Counter
5- Adder
6- Multiplexer
Solution
1. The D flip-flop tracks the input, making transitions with match those of the input D. The D
stands for \"data\"; this flip-flop stores the value that is on the data line. It can be thought of as a
basic memory cell. The Q output always takes on the state of the D input at the moment of a
rising clock edge. (or falling edge if the clock input is active low).
2. A register is a group of binary cells suitable for holding binary information. A group of
cascaded flip flops used to store related bits of information is known as a register.
4. A counter is a device which stores (and sometimes displays) the number of times a particular
event or process has occurred, often in relationship to a clock signal. Each pulse applied to the
clock input increments or decrements the number in the counter.
5. An adder, also called summer, is a digital circuit that performs addition of n-bit binary
numbers.
6. A multiplexer (or mux) is a device that selects one of several analog or digital input signals
and forwards the selected input into a single line. A multiplexer of 2n inputs has n select lines,
which are used to select which input line to send to the output..
Two sources of radio waves (call them 1 and 2) are separated by a dis.pdf
Two sources of radio waves (call them 1 and 2) are separated by a distance d along the a-axis in
the plane, and emit sinusoidal waves of the same wavelength A and frequency f in the N (+y), S
(-y), E (+x), and W (-x) directions. Suppose that we want to arrange the sources to broadcast a
signal in the E direction, but not the W. Can this be accomplished? If so, how? In not, why?
Solution
both the sources are identicle and emit in all direction N,S,E,W. by introducing a path difference
equal to half wave length the waves will have distructive interference and signal is destroyed but
this happens in both the direction E nd W equally.
Hence it is not possible to arrnge the sources such that we have signal only in E nd not in W, the
only way to achieve this is to shield the source in W direction or put reflectors in W direction so
that all W signal is reflected back to E.
Use the information in the table to select the image that represents .pdf
Use the information in the table to select the image that represents the structure of B-form DNA.
Solution
First one represents the B DNA
The structure identified by Watson and Crick, is known as B DNA and is believed to be the
idealised form of the structure adopted by virtually all DNA in vivo. It is characterized by a
helical repeat of 10 bp/turn by the presence of base pairs lying on the helix axis and almost
perpendicular to it, and by having well defined, deep major and minor grooves..
Triggers can only be used to update table values. PLSQL blocks have.pdf
Triggers can only be used to update table values. PL/SQL blocks have a section used to declare
variables. True False
Solution
Please find the answers below:
1) False; triggers are not only used to update the table values, but also they are being used in
many places like: generating derived column, enforcing referential integrity, imposing security
authorizations, etc...
2) True; PL/SQL has a dedicated section where the variables are being declared. The blocks
name is DECLARE block, which comes before the actual code of statements which is defined in
a BEGIN - END blocks..
TB is a major worldwide disease with new cases arising at an alarmin.pdf
TB is a major worldwide disease with new cases arising at an alarming rate, What factors can
account for this and what lays behind the trend of multidrug resistant strains of mycobacterium?
please cite reference.
Solution
Tuberculosis is caused by mycobacteriam tuberculosis. TB is one of the most common cause of
death due to infection of the mycobacteriam species.various factor have been found to develop
resistance against M. tuberculosis. resistance means simply afterwards that drug potency in terms
of pharmacological response will be decresed gradually and at certain level it will not active
against the strain. There are two types of resistance 1. natural resistance and second is aquired
resistance. natural resistance is found on wild strain. multidrug resistance is the phase in which
strain is not responsive against primary line therapy of disease.
factor which induced resistance
gene factor
In a recent year study suggest that the patient who shows the presence of HLA -DRB1 13 and
HLA-DRB 1 14 has two fold risk to develop resistance against strain.(Park et al.)
Incomplete and inadequate mangment Review of literature suggests that the most powerful
factor of MDR-TB is a history of treatment of tuberculosis(diagnosis). TB patients get treated
with DOTS regimens not only through the Revised National Tuberculosis Control Programme
(RNTCP), but also receive treatment from private medical doctors. Irregular, incomplete,
inadequate treatment is the commonest mean of acquiring drug resistant due to improper
mangment of drug mostly organism develop the resistnacce against drug eg. use of single drug to
treat tb is the fetal cause to develope resistance against organisms.it may be due to ignorance.
anothe common error in prescription error it may like additions syndromes.
Inadequate treatment compliance : the change over form has complied a significant effece in
development of drug resistance it was observed that only 44 per cent of the patients receiving
short-course treatment (n=2300) and 36 per cent of those receiving standard chemotherapy
(n=1000) completed 81 per cent or more of their treatment.
Noncompliance with prescribed treatment is often underestimated by the physician and is
difficult to predict. On the other hand, certain factors such as psychiatric illness, alcoholism, drug
addiction and homelessness do predict noncompliance .The drug defaulter, just like placebo
reactor is not a consistent or readily identified person demographic factors such as ,sex, age,
marital status,socio-economic status and education levelhave not been found to correlate with the
degree of compliance..
7. How did printing technology impact literacy and educationSol.pdf
7. How did printing technology impact literacy and education?
Solution
7. How did printing technology impact literacy and education?
Printing technology impacted literacy and education. Printing transformend the world and made
very important changes.
Previously people don\'t have much access to information. They used to write. So, the books are
expensive. In 15th century Johannes Gutenberg invented a printing machine and made it
available to world.
Not all are rich enough to buy hand written books till 15th century. As books are expensive so
does literacy. To learn to read and write, need to refer to books. When this printing machine
came to availability, books became cheap and affordable. Then on wards the barrier of high cost
had gone.
Source of knowledge are hand written papers or books besides people. Information is not so
much available. Again because of expensive things. When printing is available news papers
came into picture and from then on people had one more source of knowledge which is cheaper.
In this way printing technology impact literacy and education..
Question 1Theories of Law Subject__________ is a collaborative.pdf
Question 1////Theories of Law Subject
__________ is a collaborative effort of two or more agencies that provides resources, expertise,
and information with the goal of maximizing its capabilities to detect, prevent and respond to
terrorist activities.
State patrol
Federal investigations units
Fusion centers
DHS
2 points
Question 2
By 2008, the TSA had supplied about 43,000 __________ in an effort to secure air transportation
systems.
police officers
U.S. coast guard agents
border patrol agents
security personnel
2 points
Question 3
The average life expectancy of a police officer is __________ years old.
62
49
57
40
2 points
Question 4
The Department of Justice oversees all of the following agencies EXCEPT:
the Federal Bureau of Investigation.
the U.S. Marshals Service.
the Drug Enforcement Administration.
the Financial Crimes Network.
2 points
Question 5
The most controversial method of ending job discrimination is the concept of __________.
affirmative action
reverse discrimination
adverse impact
double marginality
2 points
Question 6
On average, about __________of full-time sworn personnel in a local police department are
female.
88.7%
10.2%
2.7%
11.3%
2 points
Question 7
The police attempt to prevent crime by trying to create a sense of __________.
omniscience
omnipresence
patrolling
fear
2 points
Question 8
Klockars defines the basic functions of policing as:
dealing with all those problems that may require the use of coercive force.
patrolling on foot throughout local neighborhoods.
arresting persons for crimes of vandalism.
arresting persons for violent offenses.
2 points
Question 9
The position of the Patrol Sergeant falls under the __________ in the chain of command from
Chief to Patrol officer.
Patrol Lieutenant
Operations Captain
Chief of Police
First Lieutenant
2 points
Question 10
The Supreme Court ruled that police may stop automobiles based on reasonable suspicion in all
of the following ways EXCEPT:
a crime was being committed.
a traffic violation occurred.
a random routine stop.
all citizens are subject to the stop.
2 points
Question 11
The __________ is considered one of the most important units in investigating terrorism in the
U.S.
FBI-Joint Terrorist Task Force
ATF
Department of Homeland Security
National Counterterrorism Center
2 points
Question 12
Studies have indicated that laws establishing administrative license revocations have helped
reduce alcohol-related crashes by __________ percent.
30
40
20
60
2 points
Question 13
Which of the following is NOT one of the FBI\'s post-9/11 priorities?
To protect civil rights
To combat terrorism against the United States
To protect the U.S. against cyber-based attacks and high-technology crimes
To monitor the flow of trade and travel
2 points
Question 14
__________ is the process of running an organization so that the organization can accomplish its
goals.
Bureaucracy
Management
Structure
Coordinating
2 points
Question 15
In the early days of policing in England, the watchmen in a.
Q Describe the 2 types of homologs. (1) Orthologs (2) ParalogsS.pdf
Q: Describe the 2 types of homologs. (1) Orthologs (2) Paralogs
Solution
1. Orthologs: These are genes found in different species that resulted from speciation of a
common ancestral gene. During this event of evolution, the orthologs retain their original
function.
2. Paralogs: These genes are resulted from duplication event with a genome. And during the
evolution, paralogs evolve to attain new function unlike orthologs..
Please fix the java code (using eclipse)package hw4p1;import jav.pdf
Please fix the java code (using eclipse)
package hw4p1;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
public class MeanMedianHWv2 {
public static void main(String[] args) throws IOException {
String[] arrDate = { \"\", \"\", \"\",\"\" };
Scanner in = new Scanner(System.in);
Boolean day = true;
// Year
do
{
System.out.println(\"Enter a year (2016)\");
arrDate[0] = in.next();
if (!arrDate[0].equalsIgnoreCase(\"2016\"))
{
System.out.println(\"Wrong Input. \'2016\' only! Try again.\");
}
else
{
arrDate[0] = \"2016\";
}
} while(!arrDate[0].equalsIgnoreCase(\"2016\"));
// Month
do
{
System.out.println(\"Enter a month (01)\");
arrDate[1] = in.next();
if (!arrDate[1].equalsIgnoreCase(\"01\"))
{
System.out.println(\"Wrong Input. \'01\' only! Try again!\");
}
else
{
arrDate[1] = \"01\";
}
} while(!arrDate[1].equalsIgnoreCase(\"01\"));
// Day
do
{
System.out.println(\"Enter a Day (04-10)\");
arrDate[2] = in.next();
//if (arrDate[2] > 10 || arrDate[2] < 4)
if (!arrDate[2].equalsIgnoreCase(\"04\"))
{
System.out.println(\"Wrong Inpit. \'04\' - \'10\' only! Try again!\");
}
//else if (arrDate[2] == 4)
else if (!arrDate[2].equalsIgnoreCase(\"05\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"06\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"07\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"08\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"09\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"10\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else
{
arrDate[2] = \"04\";
day = false;
arrDate[2] = \"05\";
day = false;
arrDate[2] = \"06\";
day = false;
arrDate[2] = \"07\";
day = false;
arrDate[2] = \"08\";
day = false;
arrDate[2] = \"09\";
day = false;
arrDate[2] = \"10\";
day = false;
}
} while (day);
do {
System.out.println(\"which data? \'w\' for wind, \'b\' for baro pressure, \'t\' for temp\");
arrDate[3] = in.next();
if (!arrDate[3]. equalsIgnoreCase (\"w\") && !arrDate[3].equalsIgnoreCase(\"b\") &&
!arrDate[3].equalsIgnoreCase(\"t\"))
{
System.out.println(\"Wrong input. \'w\' / \'b\' / \'t\' only! Try again\");
}
//while(!arrDate[2].equalsIgnoreCase(\"04\") && !arrDate[2].equalsIgnoreCase(\"05\") &&
!arrDate[2].equalsIgnoreCase(\"05\"));
if (arrDate[3].equalsIgnoreCase(\"w\"))
{
System.out.println(\"w\");
}
else if (arrDate[3].equalsIgnoreCase(\"b\"))
{
System.out.println(\"b\");
}
else if (arrDate[3].equalsIgnoreCase(\"t\"))
{
System.out.println(\"t\");
}
} while (!arrDate[3].equalsIgnoreCase(\"w\") && !arrDate[3].equalsIgnoreCase(\"b\") &&
!arrDate[3].equalsIgnoreCase(\"t\"));
in.close();
String Date = arrDate[0] +.
Overview You are tasked with writing a program called Social Security.pdf
Overview You are tasked with writing a program called Social Security that prompts the user to
enter the data (name and date-ofbirth)for a collection of people. The program will then process
this collection (array) of People (objects we create in another file) to determine who qualifies for
Social Security benefits. Specifications For our Social Security program we will need two java
files 1. Person java The person class which stores the data for a specific individual 2.
SocialSecurity.java (The driver) Details for these two class files appear below. Person ava UML
diagram name: String birthdate: LocalDate Person0 Person (newName: String, month: int, day:
int, year: int) NetName newName: String) void getName(): String set BirthDate month int, day:
int, year int) void get BirthDate 0 LocalDate toString(): String calculate Age(myBirth
LocalDate) int Person ava method description Two constructors 1. A Default constructor that
creates a Person whose name and birthdate are set to null. 2. A parameterized constructor that
accepts a String representation of the name, and three integers corresponding to the month, day,
and year in which the person was born. Setters and getters for the two instance variables. A
toString0 method that prints the person in the form dob
Solution
import java.time.LocalDate;
import java.time.Period;
import java.time.format.DateTimeFormatter;
import java.util.Scanner;
class Person {
String name;
LocalDate birthdate;
public Person() {
super();
}
public Person(String name, int month, int day, int year) {
super();
this.name = name;
this.birthdate = LocalDate.of(year, month, day);
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public LocalDate getBirthdate() {
return birthdate;
}
public void setBirthdate(int month, int day, int year) {
this.birthdate = LocalDate.of(year, month, day);
}
@Override
public String toString() {
DateTimeFormatter formatter = DateTimeFormatter.ofPattern(\"MM/dd/yyyy\");
birthdate.format(formatter);
return \"Person [name=\" + name + \", birthdate=\" + birthdate + \"]\";
}
public int calculateAge() {
return Period.between(birthdate, LocalDate.now()).getYears();
}
public void qualifySSN() {
if(this.calculateAge()>66)
System.out.println(\"Qualifies for Social Security Benifits\");
else
System.out.println(\"Doesnot qualifies for Social Security Benefits\");
}
}
public class SocialSecurity {
public static void main(String[] args) {
if (args.length > 1)
System.out.println(\"Invalid number of command line arguments\");
int num = Integer.parseInt(args[0]);
Person[] p = new Person[num];
Scanner in = new Scanner(System.in).useDelimiter(\"\\\ \");
;
for (int i = 0; i < num; i++) {
System.out.println(\"enter persons name\");
String name = in.next();
System.out.println(\"enter date of birth in the format Month Day Year\");
String dob = in.next();
String[] temp = dob.split(\" \");
p[i] = new Person(name, Integer.parseInt(temp[0]), Integer.parseInt(temp[1]),
Integer.pa.
A geneticist Dr. O. Sophila discovers a new mutation that causes the .pdf
A geneticist Dr. O. Sophila discovers a new mutation that causes the flies to shake and quiver.
She calls this mutation spastic (sps) and determines that spastic is due to an autosomal recessive
mutation. She wants to determine if the spastic gene is linked to the gene for vestigial wings (vg)
- also a recessive phenotype. She has true breeding stocks for each kind of mutations. She
crosses a spastic fly with a vestigial wing fly and then uses the resulting F1 females in a
testcross. She obtains the following F2 flies from the testcross: Spastic with normal wings 230
NonSpastic with vestigial wings 224 Spastic with vestigial wings 97 NonSpastic with normal
wings 99 Are the genes that cause vestigial wings and the spastic mutation linked? Provide the
rationale to your answer. What is the genotype and phenotype of F1 females? Annotate the
genotype notation illustrates whether sps and vg alleles are in cis or trans arrangement to each
other. What is(are) phenotype(s) of th
Solution
Answer:
3.1)
Spastic (recessive) = s; Nonspastic (dominant)=s+
Vestigial wings (recessive)= v; Normal wings(dominant) = V
ssVV (spastic fly) x s+ s+ vv (Vestigial wings)-----P
sV s+v----------------------------gametes
s+sVv (nonspastic, normal wing)---------F1
F1 X ssvv------------Test cross.
During test cross, all the progeny should be in equal proportion, but it is not the case. So the
genes were linked.
3.2).
F1 phenotype= nonspastic, normal wing
Genotyope: s+sVv
Recessive allele of one gene and dominant allele of the other gene present on the same
chromosome, so the configuration of the genes is trans configuration.
3.3).
The recombinant phenotypes of F2 files are= spastic with vestigial wings and nonspastic with
normal wings
3.4).
spastic with normal wings = ss Vv
Nonspastic with vestigial wings = s+svv
Spastic with vestigial wings = ssvv
Nonspastic with normal wings= s+sVv
3.5).
%Recombinant frequency = map distance (map units)
Recombinat frequency = (total number of recombinants / Total progeny)100
= 97+99 = (196/650)100 = 30.15%
The genetic distance between the v and s genes is 30.15 map units.
o Dfe the cefcien of hermal expansion. What is the relation between t.pdf
o Dfe the cefcien of hermal expansion. What is the relation between the linear and the
volumetric coefficients of thermal expansion?
Solution
is the ratio of the fractional change in size of a material to its change in temperature; is
represented by the symbol (alpha) for solids and (beta) for liquids.
relation between linear and volumetric coefficients of thermal expansion is that V=3L
Where, V = volumetric coefficients of thermal expansion
L = linear coefficients of thermal expansion.
Monohybrid crosses with incomplete dominance and co-dominance The red.pdf
Monohybrid crosses with incomplete dominance and co-dominance The red and white alleles
for the snapdragon flower color gene exhibit incomplete dominance. When pink flowers were
produced from red- and white-flowered parents, early geneticists thought this might represent
\"blending inheritance, \" where the hereditary information from one parent melded with the
information from the other. Use Punnett Squares to determine the phenotype ratios in snapdragon
offspring from the following crosses, that show that blending inheritance is not an accurate way
to characterize this situation. define the notation (letters) you will use: pink x pink (if blending
inheritance were accurate, we would predict that all offspring would be pink) pink x red (if
blending inheritance were accurate, we would predict that all offspring would be dark pink)
Solution
A cross between pink and pink flower plants.
The genotype of pink flower plants would be Rr (R allele for red colour and r allele for white
colour). A case of incomplete dominance.
The gamete formation results in R and r genotypes
And the offsprings will be red ,pink, white in the ratio of 1:2:1 respectively.
All offsprings are not pink. There is segregation of alleles during gamete formation, that results
in new combination of genes in tha offsprings. Here alleles for red and white colour were present
in the parents and their offsprings also contains new combination of alleles.
A cross between red and pink flower plants.
The genotype red is RR and pink is Rr. The cross between the two results in red and pink flowere
in equal ratio 2:2.
Here the alleles were present for red and white colour in the parents RR X Rr.
There is no occurance of blending inheritance, Here the concerned genes for red (RR) and white
(rr) colour show incompletee dominance when come together in an organism and shows pink
colour (Rr). This does not mean that it happens in all conditions, as the conditions are
exceptional cases. Here the heterozygote (Rr) posses genes for both red and white colour and
when self crossed results in new combination of genes due to their segregation during gamete
formation and homozygous condition will be retained in some plants (RR and rr) that results in
red and white flowers.RrRRRRrrRrrr.
Match the tissues in column A with the characteristics in (Some answe.pdf
Match the tissues in column A with the characteristics in (Some answers may be used more than
once.) a. Simple columnar epithelium b. Simple cuboidal epithelium c. Simple squamous
epithelium d. Pscudostratifled columnar epithelium e. Stratified squamous epithelium f.
Stratified columnar epithelium g. Stratified cuboidal epithelium h. Transitional epithelium 1.
Consists of several layers of cube-shaped, elongated, and irregular cells, allowing an expandable
lining Commonly possesses cilia that move dust and mucus out of the respiratory airways
Single layer of flattened cells Single row of elongated cells, but some cells don\'t reach the free
surface Forms walls of capillaries and air sacs of lungs Provides lining of urethra of males and
parts of pharynx Provides abrasion protection of skin epidermis and oral cavity Forms inner
lining of urinary bladder and ureters 9. Lines kidney tubules and ducts of salivary glands 10.
Forms lining of stomach and intestines 11. Two or three layers of cube-shaped cells 12. Forms
lining of oral cavity, anal canal, and vagina
Solution
A. Simple columnar epithelium- forms single layer of flattened cells, forms the inner lining of
stomach and intestines.
B. Simple cuboidal epithelium- Lines kidney tubules and ducts of salivary glands- found in area
where secretions and absorption occurs
C. Simple squamous epithelium- single layer of flattened cells, 5. Forms walls of capillaries and
air sacs of lungs
D.Pseudostratified columnar epithelium- commonly possesses cilia that move dust and mucus
out of the respiratory airway , 4. Single raw of elongated cells but some don\'t reach the free
surface.
E. Stratified squamous epithelium- 12. Forms lining of oral cavity, anal canal and vagina7.
Provides abrasion protection of skin epidermis and oral cavity.
F. Stratified columnar epithelium- 6. Provides lining of urethra and pharynx
G. Stratified cuboidal epithelium- 11. Two or three layers of cube shaped cells.
H. Transitional epithelium- 1. Consists of several layers of cube shaped, elongated and irregular
cells, allowing an expandable lining..
Jumbo shrimp are those that require 10 to 15 shrimp to make a pound..pdf
Jumbo shrimp are those that require 10 to 15 shrimp to make a pound. Suppose that the number
of jumbo shrimp in a 1-pound bag averages ? = 12.5 with a standard deviation of ? =1, and forms
a normal distribution. What is the probability of randomly picking a sample on n=25 1-pound
bags that average more than M=13 shrimp per bag?
Solution
P(xbar>13) = P((xbar-mean)/(s/vn) >(13-12.5)/(1/5))
=P(Z>2.5) = 0.0062 (from standard normal table).
In metaphase 1 of meiosis, the homologous chromosomes line up side by.pdf
In metaphase 1 of meiosis, the homologous chromosomes line up side by side along the equator
so that (a) crossing-over (exchange of DNA regions) can occur between the homologous pairs
and (b) the homologous chromosomes can be pulled to opposite poles during anaphase 1. In
mitosis, by contrast, homologous chromosomes line up single-file along the equator. What
benefits are derived from these two different ways that homologous chromosomes are positioned
at metaphase in meiosis and mitosis?
Solution
The equatorial arrangement of chromosomes make the cell divisions to occur more
symmetrically. This chromosomal alignment is the keyspot of metazoan cell divisions. The
spindle fibers in both mitosis and meiosis relocate the metaphase plates if they are not properly
arranged. Homologous chromosomes functions differently in both mitosis and meiosis. The
homologous chromosomes in mitosis do not pair with eachother like they do in meiosis. They
line up independently at the equatorial plate whereas in meiosis they occur in pairs. The chances
of chromosomes to cross one another by mitotic recombination is very rare..
Implement the sequence class from Section 3.2 of the textbook. The d.pdf
Implement the sequence class from Section 3.2 of the textbook. The documentation and
definition portion of this sequence ADT is provided in textbook (pp.124-133), and is posted in
BlackBoard of this class too.
Also implement two operators + and += for the ADT. For the + operator, x+y (a new sequence)
contains all the items of x, followed by all the items of Y.
The statement x += y appends all of the items of y to the end of what’s already in X.
For simplicity, let’s assume the sequence contains integer items and the capacity is 30.
Write a simple application to manage a playlist of your favorite music, where the sequence class
is used to insert, delete, search and print the title of music.
Your application should read in integral numbers (an ID, instead of title, for simplicity) of each
piece of music and insert them onto a sequence. It should be able to insert an item at a specified
location, find and display the music currently at play, delete a specified item from the sequence
as well as display the entire list of music in the sequence.
// FILE: sequence1.h
// CLASS PROVIDED: sequence (part of the namespace main_savitch_3)
// There is no implementation file provided for this class since it is
// an exercise from Section 3.2 of \"Data Structures and Other Objects Using C++\"
//
// TYPEDEFS and MEMBER CONSTANTS for the sequence class:
// typedef ____ value_type
// sequence::value_type is the data type of the items in the sequence. It
// may be any of the C++ built-in types (int, char, etc.), or a class with a
// default constructor, an assignment operator, and a copy constructor.
//
// typedef ____ size_type
// sequence::size_type is the data type of any variable that keeps track of
// how many items are in a sequence.
//
// static const size_type CAPACITY = _____
// sequence::CAPACITY is the maximum number of items that a sequence can hold.
//
// CONSTRUCTOR for the sequence class:
// sequence( )
// Postcondition: The sequence has been initialized as an empty sequence.
//
// MODIFICATION MEMBER FUNCTIONS for the sequence class:
// void start( )
// Postcondition: The first item on the sequence becomes the current item
// (but if the sequence is empty, then there is no current item).
//
// void advance( )
// Precondition: is_item returns true.
// Postcondition: If the current item was already the last item in the
// sequence, then there is no longer any current item. Otherwise, the new
// current item is the item immediately after the original current item.
//
// void insert(const value_type& entry)
// Precondition: size( ) < CAPACITY.
// Postcondition: A new copy of entry has been inserted in the sequence
// before the current item. If there was no current item, then the new entry
// has been inserted at the front of the sequence. In either case, the newly
// inserted item is now the current item of the sequence.
//
// void attach(const value_type& entry)
// Precondition: size( ) < CAPACITY.
// Postcondition: A new copy of entry has been ins.
Gene frequencies for the ABO alleles vary geographically. For exa.pdf
Gene frequencies for the ABO alleles vary geographically. For example the I^A allele makes up
nearly 50% of the genes for blood type among Australian Aborigines, about 25% in Central
Asian populations, and 0% in peoples native to South America. No blood type has sufficient
selective advantage over another to explain these regional differences. From these facts, what
can you infer about the history of our species Did humanity develop as a large and continuous
population spreading out across the planet, or did we disperse while still living in small scattered
groups
Solution
Based on the above data human population developed as small, some of them scattered into
small groups and acclimatised to new geographical area. In each group some dominant blood
alleles become predominant and become major allele in that population during course of time.
This is the reason for differences in gene frequencies.
How to Setup Warehouse & Location in Odoo 17 Inventory
In this slide, we'll explore how to set up warehouses and locations in Odoo 17 Inventory. This will help us manage our stock effectively, track inventory levels, and streamline warehouse operations.
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
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Which of the following correctly states the functions of the roug.pdf
Which of the following correctly states the functions of the rough endoplasmic reticulum ? lipid
metabolism, cholesterol synthesis, and synthesis of the lipid components of lipoproteins
absorption, synthesis, and transport of fats. storage of calcium ions synthesis of the cell\'s
membranes
Solution
As per the given options rough endoplasmic reticulum functions are :-
2) Absorbtion, synthesis and transport of fats & 4) Synthesis of cell membranes
Generally, the eucaryotic cell contains two types of endoplasmic reticulum
i) Rough Endoplasmic Reticulum.
Rough Endoplasmic Reticulum why named as rough means it contains ribosomes on it so, the
surface of the reticulum looks rough. The main function of the rough ER is protein synthesis due
to the ribosomes.
Rough ER through the membrane contact sites contact with the other organells allowing the
transfer of lipids and other small molecules.
The intestinal cells, pancreatic cells, which involves in the absorption process contains more
rough ER why because the proteins synthesized by the Rough ER involved in the intracellular
transport of resynthesized lipids.So the rough ER indirectly helps in the absorption process.
ii)Smooth Endoplasmic Reticulum:-
Due to the absence of ribosomes it looks smooth so named as smooth ER. The main role is lipid
metabolism, and synthesis of carbohydrates and steroid harmones and storage of calcium ions
So, in the given question the first option(Lipid metabolism, cholesterol synthesis, synthesis of
lipid components of lipo proteins) indicates smooth Endoplasmic Reticulum functions
2nd option Absorbtion, synthesis and transport of fats indicates Rough Endoplasmic Reticulum
functions
3rd Option Indicates(Storage of Calcium ions) smooth Endoplasmic Reticulum functions
4th Option is synthesis of cell membrane, For this function the main role of rough Endoplasmic
reticulum is during the lipid transport the replenishment of new membranes is done by the
morphological change of rough endoplasmic reticulum involved..
What is isomerism How many kinds are there Define each kind and.pdf
What is isomerism? How many kinds are there? Define each kind and explain with examples?
What is isomerism? How many kinds are there? Define each kind and explain with examples?
Solution
The process of isomerism involves the isomers .These are basically defined as the molecules
which have the same molecular formula but they usually possess a different arrangement of the
atoms in space. They are of two types:
1.Constitutional isomers :They are defined as the compounds in which the various atoms are
connected or bonded in a different manner .They are also called as structural isomers
2.Stereoisomers:On the other hand ,they are the compounds which differ on the basis of
orientation in space.They are of two types
a)Enantiomers :They are the compounds which are usually a mirror image of one another and
thehese mirror images are also non-superimposable
b)Diastereomers :They are isomers which are not mirror images of one another.Moreover these
compounds are non-superimposable. They are of two types: cis and trans
->Cis-isomers are the isomers which usually contain atoms which are oriented on the same side
of a bond.
->On the other hand the Trans-isomers are the compounds that have atoms oriented on opposite
sides of a bond.
What is the definition of the words, and state the diffe.pdf
What is the definition of the words, and state the difference between them if there is:
1- D-Flip Flop
2- Register
3- Equality
4- Counter
5- Adder
6- Multiplexer
What is the definition of the words, and state the difference between them if there is:
1- D-Flip Flop
2- Register
3- Equality
4- Counter
5- Adder
6- Multiplexer
What is the definition of the words, and state the difference between them if there is:
1- D-Flip Flop
2- Register
3- Equality
4- Counter
5- Adder
6- Multiplexer
Solution
1. The D flip-flop tracks the input, making transitions with match those of the input D. The D
stands for \"data\"; this flip-flop stores the value that is on the data line. It can be thought of as a
basic memory cell. The Q output always takes on the state of the D input at the moment of a
rising clock edge. (or falling edge if the clock input is active low).
2. A register is a group of binary cells suitable for holding binary information. A group of
cascaded flip flops used to store related bits of information is known as a register.
4. A counter is a device which stores (and sometimes displays) the number of times a particular
event or process has occurred, often in relationship to a clock signal. Each pulse applied to the
clock input increments or decrements the number in the counter.
5. An adder, also called summer, is a digital circuit that performs addition of n-bit binary
numbers.
6. A multiplexer (or mux) is a device that selects one of several analog or digital input signals
and forwards the selected input into a single line. A multiplexer of 2n inputs has n select lines,
which are used to select which input line to send to the output..
Two sources of radio waves (call them 1 and 2) are separated by a dis.pdf
Two sources of radio waves (call them 1 and 2) are separated by a distance d along the a-axis in
the plane, and emit sinusoidal waves of the same wavelength A and frequency f in the N (+y), S
(-y), E (+x), and W (-x) directions. Suppose that we want to arrange the sources to broadcast a
signal in the E direction, but not the W. Can this be accomplished? If so, how? In not, why?
Solution
both the sources are identicle and emit in all direction N,S,E,W. by introducing a path difference
equal to half wave length the waves will have distructive interference and signal is destroyed but
this happens in both the direction E nd W equally.
Hence it is not possible to arrnge the sources such that we have signal only in E nd not in W, the
only way to achieve this is to shield the source in W direction or put reflectors in W direction so
that all W signal is reflected back to E.
Use the information in the table to select the image that represents .pdf
Use the information in the table to select the image that represents the structure of B-form DNA.
Solution
First one represents the B DNA
The structure identified by Watson and Crick, is known as B DNA and is believed to be the
idealised form of the structure adopted by virtually all DNA in vivo. It is characterized by a
helical repeat of 10 bp/turn by the presence of base pairs lying on the helix axis and almost
perpendicular to it, and by having well defined, deep major and minor grooves..
Triggers can only be used to update table values. PLSQL blocks have.pdf
Triggers can only be used to update table values. PL/SQL blocks have a section used to declare
variables. True False
Solution
Please find the answers below:
1) False; triggers are not only used to update the table values, but also they are being used in
many places like: generating derived column, enforcing referential integrity, imposing security
authorizations, etc...
2) True; PL/SQL has a dedicated section where the variables are being declared. The blocks
name is DECLARE block, which comes before the actual code of statements which is defined in
a BEGIN - END blocks..
TB is a major worldwide disease with new cases arising at an alarmin.pdf
TB is a major worldwide disease with new cases arising at an alarming rate, What factors can
account for this and what lays behind the trend of multidrug resistant strains of mycobacterium?
please cite reference.
Solution
Tuberculosis is caused by mycobacteriam tuberculosis. TB is one of the most common cause of
death due to infection of the mycobacteriam species.various factor have been found to develop
resistance against M. tuberculosis. resistance means simply afterwards that drug potency in terms
of pharmacological response will be decresed gradually and at certain level it will not active
against the strain. There are two types of resistance 1. natural resistance and second is aquired
resistance. natural resistance is found on wild strain. multidrug resistance is the phase in which
strain is not responsive against primary line therapy of disease.
factor which induced resistance
gene factor
In a recent year study suggest that the patient who shows the presence of HLA -DRB1 13 and
HLA-DRB 1 14 has two fold risk to develop resistance against strain.(Park et al.)
Incomplete and inadequate mangment Review of literature suggests that the most powerful
factor of MDR-TB is a history of treatment of tuberculosis(diagnosis). TB patients get treated
with DOTS regimens not only through the Revised National Tuberculosis Control Programme
(RNTCP), but also receive treatment from private medical doctors. Irregular, incomplete,
inadequate treatment is the commonest mean of acquiring drug resistant due to improper
mangment of drug mostly organism develop the resistnacce against drug eg. use of single drug to
treat tb is the fetal cause to develope resistance against organisms.it may be due to ignorance.
anothe common error in prescription error it may like additions syndromes.
Inadequate treatment compliance : the change over form has complied a significant effece in
development of drug resistance it was observed that only 44 per cent of the patients receiving
short-course treatment (n=2300) and 36 per cent of those receiving standard chemotherapy
(n=1000) completed 81 per cent or more of their treatment.
Noncompliance with prescribed treatment is often underestimated by the physician and is
difficult to predict. On the other hand, certain factors such as psychiatric illness, alcoholism, drug
addiction and homelessness do predict noncompliance .The drug defaulter, just like placebo
reactor is not a consistent or readily identified person demographic factors such as ,sex, age,
marital status,socio-economic status and education levelhave not been found to correlate with the
degree of compliance..
7. How did printing technology impact literacy and educationSol.pdf
7. How did printing technology impact literacy and education?
Solution
7. How did printing technology impact literacy and education?
Printing technology impacted literacy and education. Printing transformend the world and made
very important changes.
Previously people don\'t have much access to information. They used to write. So, the books are
expensive. In 15th century Johannes Gutenberg invented a printing machine and made it
available to world.
Not all are rich enough to buy hand written books till 15th century. As books are expensive so
does literacy. To learn to read and write, need to refer to books. When this printing machine
came to availability, books became cheap and affordable. Then on wards the barrier of high cost
had gone.
Source of knowledge are hand written papers or books besides people. Information is not so
much available. Again because of expensive things. When printing is available news papers
came into picture and from then on people had one more source of knowledge which is cheaper.
In this way printing technology impact literacy and education..
Question 1Theories of Law Subject__________ is a collaborative.pdf
Question 1////Theories of Law Subject
__________ is a collaborative effort of two or more agencies that provides resources, expertise,
and information with the goal of maximizing its capabilities to detect, prevent and respond to
terrorist activities.
State patrol
Federal investigations units
Fusion centers
DHS
2 points
Question 2
By 2008, the TSA had supplied about 43,000 __________ in an effort to secure air transportation
systems.
police officers
U.S. coast guard agents
border patrol agents
security personnel
2 points
Question 3
The average life expectancy of a police officer is __________ years old.
62
49
57
40
2 points
Question 4
The Department of Justice oversees all of the following agencies EXCEPT:
the Federal Bureau of Investigation.
the U.S. Marshals Service.
the Drug Enforcement Administration.
the Financial Crimes Network.
2 points
Question 5
The most controversial method of ending job discrimination is the concept of __________.
affirmative action
reverse discrimination
adverse impact
double marginality
2 points
Question 6
On average, about __________of full-time sworn personnel in a local police department are
female.
88.7%
10.2%
2.7%
11.3%
2 points
Question 7
The police attempt to prevent crime by trying to create a sense of __________.
omniscience
omnipresence
patrolling
fear
2 points
Question 8
Klockars defines the basic functions of policing as:
dealing with all those problems that may require the use of coercive force.
patrolling on foot throughout local neighborhoods.
arresting persons for crimes of vandalism.
arresting persons for violent offenses.
2 points
Question 9
The position of the Patrol Sergeant falls under the __________ in the chain of command from
Chief to Patrol officer.
Patrol Lieutenant
Operations Captain
Chief of Police
First Lieutenant
2 points
Question 10
The Supreme Court ruled that police may stop automobiles based on reasonable suspicion in all
of the following ways EXCEPT:
a crime was being committed.
a traffic violation occurred.
a random routine stop.
all citizens are subject to the stop.
2 points
Question 11
The __________ is considered one of the most important units in investigating terrorism in the
U.S.
FBI-Joint Terrorist Task Force
ATF
Department of Homeland Security
National Counterterrorism Center
2 points
Question 12
Studies have indicated that laws establishing administrative license revocations have helped
reduce alcohol-related crashes by __________ percent.
30
40
20
60
2 points
Question 13
Which of the following is NOT one of the FBI\'s post-9/11 priorities?
To protect civil rights
To combat terrorism against the United States
To protect the U.S. against cyber-based attacks and high-technology crimes
To monitor the flow of trade and travel
2 points
Question 14
__________ is the process of running an organization so that the organization can accomplish its
goals.
Bureaucracy
Management
Structure
Coordinating
2 points
Question 15
In the early days of policing in England, the watchmen in a.
Q Describe the 2 types of homologs. (1) Orthologs (2) ParalogsS.pdf
Q: Describe the 2 types of homologs. (1) Orthologs (2) Paralogs
Solution
1. Orthologs: These are genes found in different species that resulted from speciation of a
common ancestral gene. During this event of evolution, the orthologs retain their original
function.
2. Paralogs: These genes are resulted from duplication event with a genome. And during the
evolution, paralogs evolve to attain new function unlike orthologs..
Please fix the java code (using eclipse)package hw4p1;import jav.pdf
Please fix the java code (using eclipse)
package hw4p1;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.net.URL;
import java.util.ArrayList;
import java.util.Collections;
import java.util.Scanner;
public class MeanMedianHWv2 {
public static void main(String[] args) throws IOException {
String[] arrDate = { \"\", \"\", \"\",\"\" };
Scanner in = new Scanner(System.in);
Boolean day = true;
// Year
do
{
System.out.println(\"Enter a year (2016)\");
arrDate[0] = in.next();
if (!arrDate[0].equalsIgnoreCase(\"2016\"))
{
System.out.println(\"Wrong Input. \'2016\' only! Try again.\");
}
else
{
arrDate[0] = \"2016\";
}
} while(!arrDate[0].equalsIgnoreCase(\"2016\"));
// Month
do
{
System.out.println(\"Enter a month (01)\");
arrDate[1] = in.next();
if (!arrDate[1].equalsIgnoreCase(\"01\"))
{
System.out.println(\"Wrong Input. \'01\' only! Try again!\");
}
else
{
arrDate[1] = \"01\";
}
} while(!arrDate[1].equalsIgnoreCase(\"01\"));
// Day
do
{
System.out.println(\"Enter a Day (04-10)\");
arrDate[2] = in.next();
//if (arrDate[2] > 10 || arrDate[2] < 4)
if (!arrDate[2].equalsIgnoreCase(\"04\"))
{
System.out.println(\"Wrong Inpit. \'04\' - \'10\' only! Try again!\");
}
//else if (arrDate[2] == 4)
else if (!arrDate[2].equalsIgnoreCase(\"05\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"06\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"07\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"08\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"09\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else if (!arrDate[2].equalsIgnoreCase(\"10\"))
{
System.out.println(\"Error! Invalid day. Try again!\");
}
else
{
arrDate[2] = \"04\";
day = false;
arrDate[2] = \"05\";
day = false;
arrDate[2] = \"06\";
day = false;
arrDate[2] = \"07\";
day = false;
arrDate[2] = \"08\";
day = false;
arrDate[2] = \"09\";
day = false;
arrDate[2] = \"10\";
day = false;
}
} while (day);
do {
System.out.println(\"which data? \'w\' for wind, \'b\' for baro pressure, \'t\' for temp\");
arrDate[3] = in.next();
if (!arrDate[3]. equalsIgnoreCase (\"w\") && !arrDate[3].equalsIgnoreCase(\"b\") &&
!arrDate[3].equalsIgnoreCase(\"t\"))
{
System.out.println(\"Wrong input. \'w\' / \'b\' / \'t\' only! Try again\");
}
//while(!arrDate[2].equalsIgnoreCase(\"04\") && !arrDate[2].equalsIgnoreCase(\"05\") &&
!arrDate[2].equalsIgnoreCase(\"05\"));
if (arrDate[3].equalsIgnoreCase(\"w\"))
{
System.out.println(\"w\");
}
else if (arrDate[3].equalsIgnoreCase(\"b\"))
{
System.out.println(\"b\");
}
else if (arrDate[3].equalsIgnoreCase(\"t\"))
{
System.out.println(\"t\");
}
} while (!arrDate[3].equalsIgnoreCase(\"w\") && !arrDate[3].equalsIgnoreCase(\"b\") &&
!arrDate[3].equalsIgnoreCase(\"t\"));
in.close();
String Date = arrDate[0] +.
Overview You are tasked with writing a program called Social Security.pdf
Overview You are tasked with writing a program called Social Security that prompts the user to
enter the data (name and date-ofbirth)for a collection of people. The program will then process
this collection (array) of People (objects we create in another file) to determine who qualifies for
Social Security benefits. Specifications For our Social Security program we will need two java
files 1. Person java The person class which stores the data for a specific individual 2.
SocialSecurity.java (The driver) Details for these two class files appear below. Person ava UML
diagram name: String birthdate: LocalDate Person0 Person (newName: String, month: int, day:
int, year: int) NetName newName: String) void getName(): String set BirthDate month int, day:
int, year int) void get BirthDate 0 LocalDate toString(): String calculate Age(myBirth
LocalDate) int Person ava method description Two constructors 1. A Default constructor that
creates a Person whose name and birthdate are set to null. 2. A parameterized constructor that
accepts a String representation of the name, and three integers corresponding to the month, day,
and year in which the person was born. Setters and getters for the two instance variables. A
toString0 method that prints the person in the form dob
Solution
import java.time.LocalDate;
import java.time.Period;
import java.time.format.DateTimeFormatter;
import java.util.Scanner;
class Person {
String name;
LocalDate birthdate;
public Person() {
super();
}
public Person(String name, int month, int day, int year) {
super();
this.name = name;
this.birthdate = LocalDate.of(year, month, day);
}
public String getName() {
return name;
}
public void setName(String name) {
this.name = name;
}
public LocalDate getBirthdate() {
return birthdate;
}
public void setBirthdate(int month, int day, int year) {
this.birthdate = LocalDate.of(year, month, day);
}
@Override
public String toString() {
DateTimeFormatter formatter = DateTimeFormatter.ofPattern(\"MM/dd/yyyy\");
birthdate.format(formatter);
return \"Person [name=\" + name + \", birthdate=\" + birthdate + \"]\";
}
public int calculateAge() {
return Period.between(birthdate, LocalDate.now()).getYears();
}
public void qualifySSN() {
if(this.calculateAge()>66)
System.out.println(\"Qualifies for Social Security Benifits\");
else
System.out.println(\"Doesnot qualifies for Social Security Benefits\");
}
}
public class SocialSecurity {
public static void main(String[] args) {
if (args.length > 1)
System.out.println(\"Invalid number of command line arguments\");
int num = Integer.parseInt(args[0]);
Person[] p = new Person[num];
Scanner in = new Scanner(System.in).useDelimiter(\"\\\ \");
;
for (int i = 0; i < num; i++) {
System.out.println(\"enter persons name\");
String name = in.next();
System.out.println(\"enter date of birth in the format Month Day Year\");
String dob = in.next();
String[] temp = dob.split(\" \");
p[i] = new Person(name, Integer.parseInt(temp[0]), Integer.parseInt(temp[1]),
Integer.pa.
A geneticist Dr. O. Sophila discovers a new mutation that causes the .pdf
A geneticist Dr. O. Sophila discovers a new mutation that causes the flies to shake and quiver.
She calls this mutation spastic (sps) and determines that spastic is due to an autosomal recessive
mutation. She wants to determine if the spastic gene is linked to the gene for vestigial wings (vg)
- also a recessive phenotype. She has true breeding stocks for each kind of mutations. She
crosses a spastic fly with a vestigial wing fly and then uses the resulting F1 females in a
testcross. She obtains the following F2 flies from the testcross: Spastic with normal wings 230
NonSpastic with vestigial wings 224 Spastic with vestigial wings 97 NonSpastic with normal
wings 99 Are the genes that cause vestigial wings and the spastic mutation linked? Provide the
rationale to your answer. What is the genotype and phenotype of F1 females? Annotate the
genotype notation illustrates whether sps and vg alleles are in cis or trans arrangement to each
other. What is(are) phenotype(s) of th
Solution
Answer:
3.1)
Spastic (recessive) = s; Nonspastic (dominant)=s+
Vestigial wings (recessive)= v; Normal wings(dominant) = V
ssVV (spastic fly) x s+ s+ vv (Vestigial wings)-----P
sV s+v----------------------------gametes
s+sVv (nonspastic, normal wing)---------F1
F1 X ssvv------------Test cross.
During test cross, all the progeny should be in equal proportion, but it is not the case. So the
genes were linked.
3.2).
F1 phenotype= nonspastic, normal wing
Genotyope: s+sVv
Recessive allele of one gene and dominant allele of the other gene present on the same
chromosome, so the configuration of the genes is trans configuration.
3.3).
The recombinant phenotypes of F2 files are= spastic with vestigial wings and nonspastic with
normal wings
3.4).
spastic with normal wings = ss Vv
Nonspastic with vestigial wings = s+svv
Spastic with vestigial wings = ssvv
Nonspastic with normal wings= s+sVv
3.5).
%Recombinant frequency = map distance (map units)
Recombinat frequency = (total number of recombinants / Total progeny)100
= 97+99 = (196/650)100 = 30.15%
The genetic distance between the v and s genes is 30.15 map units.
o Dfe the cefcien of hermal expansion. What is the relation between t.pdf
o Dfe the cefcien of hermal expansion. What is the relation between the linear and the
volumetric coefficients of thermal expansion?
Solution
is the ratio of the fractional change in size of a material to its change in temperature; is
represented by the symbol (alpha) for solids and (beta) for liquids.
relation between linear and volumetric coefficients of thermal expansion is that V=3L
Where, V = volumetric coefficients of thermal expansion
L = linear coefficients of thermal expansion.
Monohybrid crosses with incomplete dominance and co-dominance The red.pdf
Monohybrid crosses with incomplete dominance and co-dominance The red and white alleles
for the snapdragon flower color gene exhibit incomplete dominance. When pink flowers were
produced from red- and white-flowered parents, early geneticists thought this might represent
\"blending inheritance, \" where the hereditary information from one parent melded with the
information from the other. Use Punnett Squares to determine the phenotype ratios in snapdragon
offspring from the following crosses, that show that blending inheritance is not an accurate way
to characterize this situation. define the notation (letters) you will use: pink x pink (if blending
inheritance were accurate, we would predict that all offspring would be pink) pink x red (if
blending inheritance were accurate, we would predict that all offspring would be dark pink)
Solution
A cross between pink and pink flower plants.
The genotype of pink flower plants would be Rr (R allele for red colour and r allele for white
colour). A case of incomplete dominance.
The gamete formation results in R and r genotypes
And the offsprings will be red ,pink, white in the ratio of 1:2:1 respectively.
All offsprings are not pink. There is segregation of alleles during gamete formation, that results
in new combination of genes in tha offsprings. Here alleles for red and white colour were present
in the parents and their offsprings also contains new combination of alleles.
A cross between red and pink flower plants.
The genotype red is RR and pink is Rr. The cross between the two results in red and pink flowere
in equal ratio 2:2.
Here the alleles were present for red and white colour in the parents RR X Rr.
There is no occurance of blending inheritance, Here the concerned genes for red (RR) and white
(rr) colour show incompletee dominance when come together in an organism and shows pink
colour (Rr). This does not mean that it happens in all conditions, as the conditions are
exceptional cases. Here the heterozygote (Rr) posses genes for both red and white colour and
when self crossed results in new combination of genes due to their segregation during gamete
formation and homozygous condition will be retained in some plants (RR and rr) that results in
red and white flowers.RrRRRRrrRrrr.
Match the tissues in column A with the characteristics in (Some answe.pdf
Match the tissues in column A with the characteristics in (Some answers may be used more than
once.) a. Simple columnar epithelium b. Simple cuboidal epithelium c. Simple squamous
epithelium d. Pscudostratifled columnar epithelium e. Stratified squamous epithelium f.
Stratified columnar epithelium g. Stratified cuboidal epithelium h. Transitional epithelium 1.
Consists of several layers of cube-shaped, elongated, and irregular cells, allowing an expandable
lining Commonly possesses cilia that move dust and mucus out of the respiratory airways
Single layer of flattened cells Single row of elongated cells, but some cells don\'t reach the free
surface Forms walls of capillaries and air sacs of lungs Provides lining of urethra of males and
parts of pharynx Provides abrasion protection of skin epidermis and oral cavity Forms inner
lining of urinary bladder and ureters 9. Lines kidney tubules and ducts of salivary glands 10.
Forms lining of stomach and intestines 11. Two or three layers of cube-shaped cells 12. Forms
lining of oral cavity, anal canal, and vagina
Solution
A. Simple columnar epithelium- forms single layer of flattened cells, forms the inner lining of
stomach and intestines.
B. Simple cuboidal epithelium- Lines kidney tubules and ducts of salivary glands- found in area
where secretions and absorption occurs
C. Simple squamous epithelium- single layer of flattened cells, 5. Forms walls of capillaries and
air sacs of lungs
D.Pseudostratified columnar epithelium- commonly possesses cilia that move dust and mucus
out of the respiratory airway , 4. Single raw of elongated cells but some don\'t reach the free
surface.
E. Stratified squamous epithelium- 12. Forms lining of oral cavity, anal canal and vagina7.
Provides abrasion protection of skin epidermis and oral cavity.
F. Stratified columnar epithelium- 6. Provides lining of urethra and pharynx
G. Stratified cuboidal epithelium- 11. Two or three layers of cube shaped cells.
H. Transitional epithelium- 1. Consists of several layers of cube shaped, elongated and irregular
cells, allowing an expandable lining..
Jumbo shrimp are those that require 10 to 15 shrimp to make a pound..pdf
Jumbo shrimp are those that require 10 to 15 shrimp to make a pound. Suppose that the number
of jumbo shrimp in a 1-pound bag averages ? = 12.5 with a standard deviation of ? =1, and forms
a normal distribution. What is the probability of randomly picking a sample on n=25 1-pound
bags that average more than M=13 shrimp per bag?
Solution
P(xbar>13) = P((xbar-mean)/(s/vn) >(13-12.5)/(1/5))
=P(Z>2.5) = 0.0062 (from standard normal table).
In metaphase 1 of meiosis, the homologous chromosomes line up side by.pdf
In metaphase 1 of meiosis, the homologous chromosomes line up side by side along the equator
so that (a) crossing-over (exchange of DNA regions) can occur between the homologous pairs
and (b) the homologous chromosomes can be pulled to opposite poles during anaphase 1. In
mitosis, by contrast, homologous chromosomes line up single-file along the equator. What
benefits are derived from these two different ways that homologous chromosomes are positioned
at metaphase in meiosis and mitosis?
Solution
The equatorial arrangement of chromosomes make the cell divisions to occur more
symmetrically. This chromosomal alignment is the keyspot of metazoan cell divisions. The
spindle fibers in both mitosis and meiosis relocate the metaphase plates if they are not properly
arranged. Homologous chromosomes functions differently in both mitosis and meiosis. The
homologous chromosomes in mitosis do not pair with eachother like they do in meiosis. They
line up independently at the equatorial plate whereas in meiosis they occur in pairs. The chances
of chromosomes to cross one another by mitotic recombination is very rare..
Implement the sequence class from Section 3.2 of the textbook. The d.pdf
Implement the sequence class from Section 3.2 of the textbook. The documentation and
definition portion of this sequence ADT is provided in textbook (pp.124-133), and is posted in
BlackBoard of this class too.
Also implement two operators + and += for the ADT. For the + operator, x+y (a new sequence)
contains all the items of x, followed by all the items of Y.
The statement x += y appends all of the items of y to the end of what’s already in X.
For simplicity, let’s assume the sequence contains integer items and the capacity is 30.
Write a simple application to manage a playlist of your favorite music, where the sequence class
is used to insert, delete, search and print the title of music.
Your application should read in integral numbers (an ID, instead of title, for simplicity) of each
piece of music and insert them onto a sequence. It should be able to insert an item at a specified
location, find and display the music currently at play, delete a specified item from the sequence
as well as display the entire list of music in the sequence.
// FILE: sequence1.h
// CLASS PROVIDED: sequence (part of the namespace main_savitch_3)
// There is no implementation file provided for this class since it is
// an exercise from Section 3.2 of \"Data Structures and Other Objects Using C++\"
//
// TYPEDEFS and MEMBER CONSTANTS for the sequence class:
// typedef ____ value_type
// sequence::value_type is the data type of the items in the sequence. It
// may be any of the C++ built-in types (int, char, etc.), or a class with a
// default constructor, an assignment operator, and a copy constructor.
//
// typedef ____ size_type
// sequence::size_type is the data type of any variable that keeps track of
// how many items are in a sequence.
//
// static const size_type CAPACITY = _____
// sequence::CAPACITY is the maximum number of items that a sequence can hold.
//
// CONSTRUCTOR for the sequence class:
// sequence( )
// Postcondition: The sequence has been initialized as an empty sequence.
//
// MODIFICATION MEMBER FUNCTIONS for the sequence class:
// void start( )
// Postcondition: The first item on the sequence becomes the current item
// (but if the sequence is empty, then there is no current item).
//
// void advance( )
// Precondition: is_item returns true.
// Postcondition: If the current item was already the last item in the
// sequence, then there is no longer any current item. Otherwise, the new
// current item is the item immediately after the original current item.
//
// void insert(const value_type& entry)
// Precondition: size( ) < CAPACITY.
// Postcondition: A new copy of entry has been inserted in the sequence
// before the current item. If there was no current item, then the new entry
// has been inserted at the front of the sequence. In either case, the newly
// inserted item is now the current item of the sequence.
//
// void attach(const value_type& entry)
// Precondition: size( ) < CAPACITY.
// Postcondition: A new copy of entry has been ins.
Gene frequencies for the ABO alleles vary geographically. For exa.pdf
Gene frequencies for the ABO alleles vary geographically. For example the I^A allele makes up
nearly 50% of the genes for blood type among Australian Aborigines, about 25% in Central
Asian populations, and 0% in peoples native to South America. No blood type has sufficient
selective advantage over another to explain these regional differences. From these facts, what
can you infer about the history of our species Did humanity develop as a large and continuous
population spreading out across the planet, or did we disperse while still living in small scattered
groups
Solution
Based on the above data human population developed as small, some of them scattered into
small groups and acclimatised to new geographical area. In each group some dominant blood
alleles become predominant and become major allele in that population during course of time.
This is the reason for differences in gene frequencies.
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Please fix the java code (using eclipse)package hw4p1;import jav.pdf
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Monohybrid crosses with incomplete dominance and co-dominance The red.pdf
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Match the tissues in column A with the characteristics in (Some answe.pdf
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In metaphase 1 of meiosis, the homologous chromosomes line up side by.pdf
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Date: May 29, 2024
Tags: Information Security, ISO/IEC 27001, ISO/IEC 42001, Artificial Intelligence, GDPR
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PCOS corelations and management through Ayurveda.
This presentation includes basic of PCOS their pathology and treatment and also Ayurveda correlation of PCOS and Ayurvedic line of treatment mentioned in classics.
BBR 2024 Summer Sessions Interview Training
Qualitative research interview training by Professor Katrina Pritchard and Dr Helen Williams
Leveraging Generative AI to Drive Nonprofit Innovation
In this webinar, participants learned how to utilize Generative AI to streamline operations and elevate member engagement. Amazon Web Service experts provided a customer specific use cases and dived into low/no-code tools that are quick and easy to deploy through Amazon Web Service (AWS.)
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আমাদের সবার জন্য খুব খুব গুরুত্বপূর্ণ একটি বই ..বিসিএস, ব্যাংক, ইউনিভার্সিটি ভর্তি ও যে কোন প্রতিযোগিতা মূলক পরীক্ষার জন্য এর খুব ইম্পরট্যান্ট একটি বিষয় ...তাছাড়া বাংলাদেশের সাম্প্রতিক যে কোন ডাটা বা তথ্য এই বইতে পাবেন ...
তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
বিসিএস ও ব্যাংক এর লিখিত পরীক্ষা ...+এছাড়া মাধ্যমিক ও উচ্চমাধ্যমিকের স্টুডেন্টদের জন্য অনেক কাজে আসবে ...
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LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
Hindi varnamala | hindi alphabet PPT.pdf
हिंदी वर्णमाला पीपीटी, hindi alphabet PPT presentation, hindi varnamala PPT, Hindi Varnamala pdf, हिंदी स्वर, हिंदी व्यंजन, sikhiye hindi varnmala, dr. mulla adam ali, hindi language and literature, hindi alphabet with drawing, hindi alphabet pdf, hindi varnamala for childrens, hindi language, hindi varnamala practice for kids, https://www.drmullaadamali.com
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