Answer Grow mutagenized yeast on rich media. Then grow them on mini.pdf
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Answer: Grow mutagenized yeast on rich media. Then grow them on minimal media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that do not grow on minimal media, do not grow on media supplemented with only histidine or leucine, but do grow on minimal media supplemented with both histidine and leucine Reason: A genetic screen to isolate nutritional mutants in yeast needed to isolate a double mutant that cannot synthesize histidine or leucine, two nutrients essential for growth. Initially a wild type yeast strain subjected to mutagen irradiation with a UV light followed by promoting growth of mutagenized yeast on rich media & followed by its growth on minimal media, as well as on minimal media containing histidine, minimal media containing leucine. This process is essential to isolating such a mutant using minimal media containing both histidine and leucin finally yeastis going to be selected that do not grow on minimal media supplemented with either histidine or leucine finally a double mutant can be isolated of yeast that do grow on minimal media supplemented with both histidine and leucine. Solution Answer: Grow mutagenized yeast on rich media. Then grow them on minimal media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that do not grow on minimal media, do not grow on media supplemented with only histidine or leucine, but do grow on minimal media supplemented with both histidine and leucine Reason: A genetic screen to isolate nutritional mutants in yeast needed to isolate a double mutant that cannot synthesize histidine or leucine, two nutrients essential for growth. Initially a wild type yeast strain subjected to mutagen irradiation with a UV light followed by promoting growth of mutagenized yeast on rich media & followed by its growth on minimal media, as well as on minimal media containing histidine, minimal media containing leucine. This process is essential to isolating such a mutant using minimal media containing both histidine and leucin finally yeastis going to be selected that do not grow on minimal media supplemented with either histidine or leucine finally a double mutant can be isolated of yeast that do grow on minimal media supplemented with both histidine and leucine..
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The Periodic table of elements are classified through color depending on whether they are
metals or non – metals,
which is determined through their physical characteristics and chemical properties.
Physically, metals in general are usually solid when in room temperature.
They are known to have high density and luster, which means they are quite heavy in nature and
interact well with light
They are very malleable, means which makes them capable of being hammered into sheets or
other shapes,
and are also ductile, which makes them able of being expanded.
Solution
The Periodic table of elements are classified through color depending on whether they are
metals or non – metals,
which is determined through their physical characteristics and chemical properties.
Physically, metals in general are usually solid when in room temperature.
They are known to have high density and luster, which means they are quite heavy in nature and
interact well with light
They are very malleable, means which makes them capable of being hammered into sheets or
other shapes,
and are also ductile, which makes them able of being expanded.
.
#include iostream #include string #include invalidHr.h.pdf
#include
#include
#include \"invalidHr.h\"
#include \"invalidMin.h\"
#include \"invalidSec.h\"
using namespace std;
int getHours();
int getMinutes();
int getSeconds();
void print24HourTime(int hr, int min, int sec, string str);
int main ()
{
int hours;
int minutes;
int seconds;
string str;
hours = getHours();
minutes = getMinutes();
seconds = getSeconds();
cout << \"Enter AM or PM: \";
cin >> str;
cout << endl;
cout << \"24 hour clock time: \";
print24HourTime(hours, minutes, seconds, str);
system(\"pause\");
return 0;
}
int getHours()
{
bool done = false;
int hr = 0;
do
{
try
{
cout << \"Enter hours: \";
cin >> hr;
cout << endl;
if (hr < 0 || hr > 12)
throw invalidHr();
done = true;
}
catch (invalidHr hrObj)
{
cout << hrObj.what() << endl;
}
}
while (!done);
return hr;
}
int getMinutes()
{
bool done = false;
int min = 0;
do
{
try
{
cout << \"Enter minutes: \";
cin >> min;
cout << endl;
if (min < 0 || min > 59)
throw invalidMin();
done = true;
}
catch (invalidMin minObj)
{
cout << minObj.what() << endl;
}
}
while (!done);
return min;
}
int getSeconds()
{
bool done = false;
int sec = 0;
do
{
try
{
cout << \"Enter seconds: \";
cin >> sec;
cout << endl;
if (sec < 0 || sec > 59)
throw invalidSec();
done = true;
}
catch (invalidSec secObj)
{
cout << secObj.what() << endl;
}
}
while (!done);
return sec;
}
void print24HourTime(int hr, int min, int sec, string str)
{
if (str == \"AM\")
{
if (hr == 12)
cout << 0;
else
cout << hr;
cout << \":\" << min << \":\" << sec << endl;
}
else if (str == \"PM\")
{
if (hr == 12)
cout << hr;
else
cout << hr + 12;
cout << \":\" << min << \":\" << sec << endl;
}
}
invalidHr.h
#include
#include
using namespace std;
class invalidHr
{
public:
invalidHr() // define message
{
message=\"The value of hr must be between 0 and 12.\";
}
invalidHr(string str)
{
message = str + \"nope\";
}
string what() // throw what message object
{
return message;
}
private:
string message;
};
invalidMin.h
#include
#include
using namespace std;
class invalidMin
{
public:
invalidMin() // define message
{
message=\"The value of min must be between 0 and 59.\";
}
invalidMin(string str)
{
message = str + \"nope\";
}
string what() // return message if what is called
{
return message;
}
private:
string message;
};
invalidSec.h
#include
#include
using namespace std;
class invalidSec // define the class
{
public:
invalidSec() // define message
{
message=\"The value of sec must be between 0 and 59\";
}
invalidSec(string str)
{
message = str + \"nope\";
}
string what() // return the message if what is called
{
return message;
}
private:
string message;
};
Solution
#include
#include
#include \"invalidHr.h\"
#include \"invalidMin.h\"
#include \"invalidSec.h\"
using namespace std;
int getHours();
int getMinutes();
int getSeconds();
void print24HourTime(int hr, int min, int sec, string str);
int main ()
{
int hours;
int minutes;
int seconds;
string str;
hours = getHours();
minutes = getMinutes();
seconds = getSeconds();
cout << \"Enter AM or PM: \";
cin >> .
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strain improvement to increase yeild of selected molecules.pptx
More from APMRETAIL
The Periodic table of elements are classified through color dependi.pdf
The Periodic table of elements are classified through color depending on whether they are
metals or non – metals,
which is determined through their physical characteristics and chemical properties.
Physically, metals in general are usually solid when in room temperature.
They are known to have high density and luster, which means they are quite heavy in nature and
interact well with light
They are very malleable, means which makes them capable of being hammered into sheets or
other shapes,
and are also ductile, which makes them able of being expanded.
Solution
The Periodic table of elements are classified through color depending on whether they are
metals or non – metals,
which is determined through their physical characteristics and chemical properties.
Physically, metals in general are usually solid when in room temperature.
They are known to have high density and luster, which means they are quite heavy in nature and
interact well with light
They are very malleable, means which makes them capable of being hammered into sheets or
other shapes,
and are also ductile, which makes them able of being expanded.
.
#include iostream #include string #include invalidHr.h.pdf
#include
#include
#include \"invalidHr.h\"
#include \"invalidMin.h\"
#include \"invalidSec.h\"
using namespace std;
int getHours();
int getMinutes();
int getSeconds();
void print24HourTime(int hr, int min, int sec, string str);
int main ()
{
int hours;
int minutes;
int seconds;
string str;
hours = getHours();
minutes = getMinutes();
seconds = getSeconds();
cout << \"Enter AM or PM: \";
cin >> str;
cout << endl;
cout << \"24 hour clock time: \";
print24HourTime(hours, minutes, seconds, str);
system(\"pause\");
return 0;
}
int getHours()
{
bool done = false;
int hr = 0;
do
{
try
{
cout << \"Enter hours: \";
cin >> hr;
cout << endl;
if (hr < 0 || hr > 12)
throw invalidHr();
done = true;
}
catch (invalidHr hrObj)
{
cout << hrObj.what() << endl;
}
}
while (!done);
return hr;
}
int getMinutes()
{
bool done = false;
int min = 0;
do
{
try
{
cout << \"Enter minutes: \";
cin >> min;
cout << endl;
if (min < 0 || min > 59)
throw invalidMin();
done = true;
}
catch (invalidMin minObj)
{
cout << minObj.what() << endl;
}
}
while (!done);
return min;
}
int getSeconds()
{
bool done = false;
int sec = 0;
do
{
try
{
cout << \"Enter seconds: \";
cin >> sec;
cout << endl;
if (sec < 0 || sec > 59)
throw invalidSec();
done = true;
}
catch (invalidSec secObj)
{
cout << secObj.what() << endl;
}
}
while (!done);
return sec;
}
void print24HourTime(int hr, int min, int sec, string str)
{
if (str == \"AM\")
{
if (hr == 12)
cout << 0;
else
cout << hr;
cout << \":\" << min << \":\" << sec << endl;
}
else if (str == \"PM\")
{
if (hr == 12)
cout << hr;
else
cout << hr + 12;
cout << \":\" << min << \":\" << sec << endl;
}
}
invalidHr.h
#include
#include
using namespace std;
class invalidHr
{
public:
invalidHr() // define message
{
message=\"The value of hr must be between 0 and 12.\";
}
invalidHr(string str)
{
message = str + \"nope\";
}
string what() // throw what message object
{
return message;
}
private:
string message;
};
invalidMin.h
#include
#include
using namespace std;
class invalidMin
{
public:
invalidMin() // define message
{
message=\"The value of min must be between 0 and 59.\";
}
invalidMin(string str)
{
message = str + \"nope\";
}
string what() // return message if what is called
{
return message;
}
private:
string message;
};
invalidSec.h
#include
#include
using namespace std;
class invalidSec // define the class
{
public:
invalidSec() // define message
{
message=\"The value of sec must be between 0 and 59\";
}
invalidSec(string str)
{
message = str + \"nope\";
}
string what() // return the message if what is called
{
return message;
}
private:
string message;
};
Solution
#include
#include
#include \"invalidHr.h\"
#include \"invalidMin.h\"
#include \"invalidSec.h\"
using namespace std;
int getHours();
int getMinutes();
int getSeconds();
void print24HourTime(int hr, int min, int sec, string str);
int main ()
{
int hours;
int minutes;
int seconds;
string str;
hours = getHours();
minutes = getMinutes();
seconds = getSeconds();
cout << \"Enter AM or PM: \";
cin >> .
option D) answer A and B are both correct .pdf
This document discusses a multiple choice question where the correct answer is option D, stating that both answers A and B are correct. The document simply repeats the question and solution without providing any additional context or details about the question topic.
image not visible clearly. please give points to .pdf
image not visible clearly. please give points to me.
Solution
image not visible clearly. please give points to me..
HI(g) has the highest molar entropy as it has h.pdf
HI(g) has the highest molar entropy as it has highest molar mass and is gas.
Solution
HI(g) has the highest molar entropy as it has highest molar mass and is gas..
ion-dipole Solution ion-.pdf
El documento trata sobre las interacciones ión-dipolo. Menciona las soluciones iónicas y las interacciones entre iones y moléculas polares.
Yes. It has SP2 hybridisation , with 2 lone pairs.pdf
Yes. It has SP2 hybridisation , with 2 lone pairs and 2 single bonds.
Solution
Yes. It has SP2 hybridisation , with 2 lone pairs and 2 single bonds..
Using, M1V1 = M2V2 10.0 x 0.1106 = M2 x 13.64 M.pdf
Using, M1V1 = M2V2 10.0 x 0.1106 = M2 x 13.64 M2 = Molarity of NaOH =
0.081 M
Solution
Using, M1V1 = M2V2 10.0 x 0.1106 = M2 x 13.64 M2 = Molarity of NaOH =
0.081 M.
The start of the European Colonization is typically dated to 1492, a.pdf
The start of the European Colonization is typically dated to 1492, although there was at least one
earlier colonization effort. The first known Europeans to reach the Americas are believed to have
been the Vikings (\"Norse\") during the eleventh century, who established several colonies in
Greenland and one short-lived settlement at L\'Anse aux Meadows in the area the Norse called
Vinland, present day Newfoundland. Settlements in Greenland survived for several centuries,
during which time the Greenland Norse and theInuit people experienced mostly hostile contact.
By the end of the fifteenth century, the Norse Greenland settlements had collapsed. In 1492, a
Spanish expedition headed by Christopher Columbus reached the Americas, after which
European exploration and colonization rapidly expanded, first through much of the Caribbean
region (including the islands of Hispaniola,Puerto Rico, and Cuba) and, early in the sixteenth
century, parts of the mainlands of North and South America.
Eventually, the entire Western Hemisphere would come under the domination of European
nations, leading to profound changes to its landscape, population, and plant and animal life. In
the nineteenth century alone over 50 million people left Europe for the Americas. The post-1492
era is known as the period of the Columbian Exchange. The potato, the pineapple, theturkey,
dahlias, sunflowers, magnolias, maize, chilies, and chocolate went East across theAtlantic
Ocean. Smallpox and measles but also the horse and the gun traveled West.
The flow of benefit appears to have been one-sided, with Europe gaining more. However, the
colonization and exploration of the Americas also transformed the world, eventually adding 31
newnation-states to the global community. On the one hand, the cultural and religious arrogance
that led settlers to deny anything of value in pre-Columbian America was destructive, even
genocidal. On the other hand, many of those who settled in the New World were also social and
political visionaries, who found opportunities there, on what for them was a tabula rasa, to aim at
achieving their highest ideals of justice, equality, and freedom. Some of the world\'s most stable
democraciesexist as a result of this transformative process.
The first conquests were made by the Spanish and the Portuguese. In the 1494 Treaty of
Tordesillas, ratified by the Pope, these two kingdoms divided the entire non-European world
between themselves, with a line drawn through South America. Based on this Treaty, and the
claims by Spanish explorer Vasco Núñez de Balboa to all lands touching the Pacific Ocean, the
Spanish rapidly conquered territory, overthrowing the Aztec and Inca Empires to gain control of
much of western South America, Central America, and Mexico by the mid-sixteenth century, in
addition to its earlier Caribbean conquests. Over this same time frame, Portugalconquered much
of eastern South America, naming it Brazil.
Early conquests, claims, and colonies
Other Eur.
The institutions which act as mediator between savers and boorrowers.pdf
The institutions which act as mediator between savers and boorrowers are known as financial
intermediateries.
The answer is d.
financial intermediaries
Solution
The institutions which act as mediator between savers and boorrowers are known as financial
intermediateries.
The answer is d.
financial intermediaries.
Solution Mitochondria is known as power house of the cell.It prod.pdf
Solution
:
Mitochondria is known as power house of the cell.It produces energy needed for different
cellular function in the body.Energy is generated in the form of ATP(Adenosine
triphosphate).Mitochondria provide all needed energy which is enough to produce our body
weight in ATP everyday.If we are more active,requirement of ATP increased.Organs like brain
and heart can not perform their function without ATP..
Sewage before being disposed of either in river stream or on land, h.pdf
Sewage before being disposed of either in river stream or on land, has generally to be treated, so
as to make it safe.
The degree of treatment required, however, depends upon the characteristics of the source of
disposal.
Sewage can be treated in different ways. treatment process are often classified as:
1) Preliminary treatment
2) Primary treatment
3) Secondary treatment
4) tertiary treatment
1) PRELIMINARY TREATMENT
It consists in solely in separating the floating materials (like dead animals, tree branches, papers,
pieces of rags, wood, etc) and also heavy settleable inorganic solids. it also helps in removing the
oil and greases etc. from the sewage.
Grit chamber or detritus tank for removing grit and sand; and skimming tanks for removing oils
and greases.
2) PRIMARY TREATMENT
Primary treatment consists in removing large suspended organic solids. this is usually
accomplished by sedimentation in settling basins.
sometimes, the preliminary as well as primary treatments are classified together under primary
treatment
the organic solids, which are seperated out in the sedimentation tanks are often stabilised by
anaerobic decomposition in a digestion tank or are incenerated. the residue for landfills or soil
conditioners
3) SECONDARY TREATMENT
Secondary treatment involves further treatment of effluents, coming from the primary
sedimentation tank. this is generally accomplished through biological decomposition of organic
matter, which can be carried out either under aerobic or anaerobic conditions. in these biological
units, bacteria will decompose the fine organic matter to produce a clearer effluent.
The treatment reactors, in which the organic matter is decomposed by aerobic bacteria are known
as aerobic biological units; and may consists of:
(i) Filters : intermittent sand filters as well as trickling filters
(ii) Aeration tank: with the feed of recycled activated sludge ( i.e. the sludge which is settled in
secondary sedimentation tank, receiving effluents from the aeration tank)
(iii) Oxidation ponds and aerated lagoons
The treatment reactors, in which the organic matter is destroyed and stabilized by anaerobic
bacteria, are known as anaerobic biological units and may consists of anaerobic lagoons, septic
tanks, imhoff tanks, etc. out of these units only anaerobic lagoons make use of primary settled
sewage and hence, they only can be classified under biological units. septic tanks andimhoff
tanks using raw sewage are therefore not classified as secondary units.
The organic solids separated out in the primary as well as in secondary settling tanks will be
disposed of by stabilizing them under anaerobic process in a sludge digestion tank.
4) FINAL or ADVANCED TREATMENT
This treatment is called as tertiary treatment. consists of removing the organic load left after the
secondary treatment and particularly to kill the pathogenic bacteria. this treatment, which is
normally carried out by chlorination, especially when treated sewage is to be di.
Relational database was proposed by Edgar Codd (of IBM Research) aro.pdf
Relational database was proposed by Edgar Codd (of IBM Research) around 1969. It has since
become the dominant database model for commercial applications (in comparison with other
database models such as hierarchical, network and object models). Today, there are many
commercial Relational Database Management System (RDBMS), such as Oracle, IBM DB2 and
Microsoft SQL Server. There are also many free and open-source RDBMS, such as MySQL,
mSQL (mini-SQL) and the embedded JavaDB (Apache Derby).
A relational database organizes data in tables (or relations). A table is made up of rows and
columns. A row is also called a record (or tuple). A column is also called a field (or attribute). A
database table is similar to a spreadsheet. However, the relationships that can be created among
the tables enable a relational database to efficiently store huge amount of data, and effectively
retrieve selected data.
A language called SQL (Structured Query Language) was developed to work with relational
databases.
Database Design Objective
A well-designed database shall:
Relational Database Design Process
Database design is more art than science, as you have to make many decisions. Databases are
usually customized to suit a particular application. No two customized applications are alike, and
hence, no two database are alike. Guidelines (usually in terms of what not to do instead of what
to do) are provided in making these design decision, but the choices ultimately rest on the you -
the designer.
Step 1: Define the Purpose of the Database (Requirement Analysis)
Gather the requirements and define the objective of your database, e.g. ...
Drafting out the sample input forms, queries and reports, often helps.
Step 2: Gather Data, Organize in tables and Specify the Primary Keys
Once you have decided on the purpose of the database, gather the data that are needed to be
stored in the database. Divide the data into subject-based tables.
Choose one column (or a few columns) as the so-called primary key, which uniquely identify the
each of the rows.
Primary Key
In the relational model, a table cannot contain duplicate rows, because that would create
ambiguities in retrieval. To ensure uniqueness, each table should have a column (or a set of
columns), called primary key, that uniquely identifies every records of the table. For example, an
unique number customerID can be used as the primary key for the Customers table; productCode
for Products table; isbn forBooks table. A primary key is called a simple key if it is a single
column; it is called a composite key if it is made up of several columns.
Most RDBMSs build an index on the primary key to facilitate fast search and retrieval.
The primary key is also used to reference other tables (to be elaborated later).
You have to decide which column(s) is to be used for primary key. The decision may not be
straight forward but the primary key shall have these properties:
Consider the followings in choose the primary key:
Let\'s il.
Option “D” is not true about the net neutrality.Favouring of net n.pdf
Option “D” is not true about the net neutrality.
Favouring of net neutrality argue that without new regulations, Internet service providers would
be able to favour their own private protocols over others
ISPS are able to encourage the use of specific services by utilising private networks to
discriminate what data is counted against bandwidth caps.
Solution
Option “D” is not true about the net neutrality.
Favouring of net neutrality argue that without new regulations, Internet service providers would
be able to favour their own private protocols over others
ISPS are able to encourage the use of specific services by utilising private networks to
discriminate what data is counted against bandwidth caps..
NaF will dissociate 100, and therefore the F- will have extra 0.25 .pdf
NaF will dissociate 100%, and therefore the F- will have extra 0.25 concentration...
HF --- > H+ + F-
[HF] = 0.3-0.3*
[H+] = 0.3*
[F-] = 0.3* + 0.25
Ka = (0.3*)*(0.3* + 0.25)/(0.3-0.3*)
7.2*10^-4 = *(0.3* + 0.25)/(1-)
= 0.0028
[H+] = 0.3*0.0028 = 0.00084 M
pH = -log(0.00084) = 3.07
Solution
NaF will dissociate 100%, and therefore the F- will have extra 0.25 concentration...
HF --- > H+ + F-
[HF] = 0.3-0.3*
[H+] = 0.3*
[F-] = 0.3* + 0.25
Ka = (0.3*)*(0.3* + 0.25)/(0.3-0.3*)
7.2*10^-4 = *(0.3* + 0.25)/(1-)
= 0.0028
[H+] = 0.3*0.0028 = 0.00084 M
pH = -log(0.00084) = 3.07.
CO2 because the heaviest would have the most attr.pdf
CO2 because the heaviest would have the most attraction ,although the deviation is
little since all have london dispersion forces between then CH4 is larger but nonpolar, hence
there will be little intermolecluar attraction. N2 is a smaller molecule,hence doesnt have much
attraction
Solution
CO2 because the heaviest would have the most attraction ,although the deviation is
little since all have london dispersion forces between then CH4 is larger but nonpolar, hence
there will be little intermolecluar attraction. N2 is a smaller molecule,hence doesnt have much
attraction.
Intensity of light source Io = 1000 countsIntensity after absorpti.pdf
Intensity of light source Io = 1000 counts
Intensity after absorption I = 392 counts
Transmittance T = (I/Io)
= 392/1000 = 0.392
Absorbance A = -log T
= -log(0.392)
= 0.407
Solution
Intensity of light source Io = 1000 counts
Intensity after absorption I = 392 counts
Transmittance T = (I/Io)
= 392/1000 = 0.392
Absorbance A = -log T
= -log(0.392)
= 0.407.
I hereby explain the SDU (service data unit )and PDU ( protocol data.pdf
I hereby explain the SDU (service data unit )and PDU ( protocol data unit) in the view of
transmitter.
Let us understnad the PDUs at different layers
Application layer - PDU is data
Presentation layer - PDU is data
Session layer - PDU is data
Transport layer - PDU segment
Network layer - PDU packet
Data link layer PDU frames
Physical layer PDU bits.
while transmission application, presentation and session layer data is incapsulated and forwarded
to Transport layer. At transport , TCP segments are created as layer 4 PDUs. When passed to IP,
they are treated as layer 3 SDUs. The IP software packages these SDUs into messages called IP
packets or IP datagrams, which are layer 3 PDUs. These are in turn passed down to a layer 2
protocol, say Ethernet, which treats IP datagrams as layer 2 SDUs, and packages them into layer
2 PDUs (Ethernet frames) which are sent on layer 1 .
On the receiving device, the process of encapsulation is reversed. The Ethernet software inspects
the layer 2 PDU (Ethernet frame) and removes from it the layer 2 SDU (IP datagram) which it
passes up to IP as a layer 3 PDU. The IP layer removes the layer 3 SDU (TCP segment) and
passes it to TCP as a layer 4 PDU.
Consclusion is the PDU from upper layer added and encapsulated at sender end and removes and
decapsulated at receiving end.
This is the way communication happens.
Hope it makes sense.
ThanksName used on OSIName Used in TCP/IP NetworksLayer 1 - PhysicalLayer 1 - Network
Access/InterfaceLayer 2 - Data linkLayer 2 - NetworkLayer 3 - NetworkLayer 3- TransportLayer
4 - TransportLayer 4 - ApplicationLayer 5 - sessionThere are only four layers defined in TCP/IP
model
Solution
I hereby explain the SDU (service data unit )and PDU ( protocol data unit) in the view of
transmitter.
Let us understnad the PDUs at different layers
Application layer - PDU is data
Presentation layer - PDU is data
Session layer - PDU is data
Transport layer - PDU segment
Network layer - PDU packet
Data link layer PDU frames
Physical layer PDU bits.
while transmission application, presentation and session layer data is incapsulated and forwarded
to Transport layer. At transport , TCP segments are created as layer 4 PDUs. When passed to IP,
they are treated as layer 3 SDUs. The IP software packages these SDUs into messages called IP
packets or IP datagrams, which are layer 3 PDUs. These are in turn passed down to a layer 2
protocol, say Ethernet, which treats IP datagrams as layer 2 SDUs, and packages them into layer
2 PDUs (Ethernet frames) which are sent on layer 1 .
On the receiving device, the process of encapsulation is reversed. The Ethernet software inspects
the layer 2 PDU (Ethernet frame) and removes from it the layer 2 SDU (IP datagram) which it
passes up to IP as a layer 3 PDU. The IP layer removes the layer 3 SDU (TCP segment) and
passes it to TCP as a layer 4 PDU.
Consclusion is the PDU from upper layer added and encapsulated at sender end and removes and
decapsulated at receiving end.
This is .
Hop1 = 13, p2 = 13, and p3 = 13Ha at least one p is not equal.pdf
Ho:p1 = 1/3, p2 = 1/3, and p3 = 1/3
Ha: at least one p is not equal to 1/3
observedexpectedO - E(O - E)
Solution
Ho:p1 = 1/3, p2 = 1/3, and p3 = 1/3
Ha: at least one p is not equal to 1/3
observedexpectedO - E(O - E).
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#include iostream #include string #include invalidHr.h.pdf
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I hereby explain the SDU (service data unit )and PDU ( protocol data.pdf
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Hop1 = 13, p2 = 13, and p3 = 13Ha at least one p is not equal.pdf
Hop1 = 13, p2 = 13, and p3 = 13Ha at least one p is not equal.pdf
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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.
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Answer Grow mutagenized yeast on rich media. Then grow them on mini.pdf
- 1. Answer: Grow mutagenized yeast on rich media. Then grow them on minimal media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that do not grow on minimal media, do not grow on media supplemented with only histidine or leucine, but do grow on minimal media supplemented with both histidine and leucine Reason: A genetic screen to isolate nutritional mutants in yeast needed to isolate a double mutant that cannot synthesize histidine or leucine, two nutrients essential for growth. Initially a wild type yeast strain subjected to mutagen irradiation with a UV light followed by promoting growth of mutagenized yeast on rich media & followed by its growth on minimal media, as well as on minimal media containing histidine, minimal media containing leucine. This process is essential to isolating such a mutant using minimal media containing both histidine and leucin finally yeastis going to be selected that do not grow on minimal media supplemented with either histidine or leucine finally a double mutant can be isolated of yeast that do grow on minimal media supplemented with both histidine and leucine. Solution Answer: Grow mutagenized yeast on rich media. Then grow them on minimal media, as well as on minimal media containing histidine, minimal media containing leucine, and minimal media containing both histidine and leucine. Select for yeast that do not grow on minimal media, do not grow on media supplemented with only histidine or leucine, but do grow on minimal media supplemented with both histidine and leucine Reason: A genetic screen to isolate nutritional mutants in yeast needed to isolate a double mutant that cannot synthesize histidine or leucine, two nutrients essential for growth. Initially a wild type yeast strain subjected to mutagen irradiation with a UV light followed by promoting growth of mutagenized yeast on rich media & followed by its growth on minimal media, as well as on minimal media containing histidine, minimal media containing leucine. This process is essential to isolating such a mutant using minimal media containing both histidine and leucin finally yeastis going to be selected that do not grow on minimal media supplemented with either histidine or leucine finally a double mutant can be isolated of yeast that do grow on minimal media supplemented with both histidine and leucine.