Option CExplanationThe word LEADING has 7 diffe.pdf
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Option C Explanation: The word \'LEADING\' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. Solution Option C Explanation: The word \'LEADING\' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720..
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2. Ligamentum arteriosumThe aortic arch loops over the left pulmo.pdf
2. Ligamentum arteriosum:
The aortic arch loops over the left pulmonary artery and the bifurcation of the pulmonary trunk,
to which it remains connected by the ligamentum arteriosum, a remnant of the fetal circulation
that is obliterated a few days after birth.
3. Caecum:
The cecum is a pouch-like dead-end passage that branches inferiorly from the end of the ileum
(end part of small intestine). Fecal matter entering the large intestine from the ileum passes into
the cecum before being pushed superiorly into the ascending colon. The appendix is attached to
the inferior end of the cecum and is believed to store beneficial bacteria that help break down
undigested food.
Solution
2. Ligamentum arteriosum:
The aortic arch loops over the left pulmonary artery and the bifurcation of the pulmonary trunk,
to which it remains connected by the ligamentum arteriosum, a remnant of the fetal circulation
that is obliterated a few days after birth.
3. Caecum:
The cecum is a pouch-like dead-end passage that branches inferiorly from the end of the ileum
(end part of small intestine). Fecal matter entering the large intestine from the ileum passes into
the cecum before being pushed superiorly into the ascending colon. The appendix is attached to
the inferior end of the cecum and is believed to store beneficial bacteria that help break down
undigested food..
moles should be equal to complete consumption so.pdf
moles should be equal to complete consumption so v*0.250 = 0.150*0.100 /2 ( 2 is
for - as Pb(NO3)2 gives 2 moles of NO3- when dissociate) so v = 0.03 L ANSWER
Solution
moles should be equal to complete consumption so v*0.250 = 0.150*0.100 /2 ( 2 is
for - as Pb(NO3)2 gives 2 moles of NO3- when dissociate) so v = 0.03 L ANSWER.
TrueOne of the fundamental rules which control the PLSS system is .pdf
True
One of the fundamental rules which control the PLSS system is that the boundaries of the public
land once approved and accepted is unchangeable.
Solution
True
One of the fundamental rules which control the PLSS system is that the boundaries of the public
land once approved and accepted is unchangeable..
The stronger the attractions betweenparticles (mo.pdf
The stronger the attractions betweenparticles (molecules or ions), the more difficult
it will be toseparate the particles. When substances melt, the particles arestill close to one
another but the forces of attraction that heldthe particles rigidly together in the solid state have
beensufficiently overcome to allow the particles to move. Whensubstances boil, the particles are
completely separated from oneanother and the attractions between molecules are
completelyovercome. The energy required to cause substances to melt and toboil, and thus
disrupt the forces of attraction, comes from theenvironment surrounding the material. If you
place a piece of icein your hand, the ice will melt more quickly than if it is placedon a cold
counter top. The energy required to melt the ice comesfrom your hand, your hand gets colder and
the ice gets warmer. for ice melting hydrogen bonding for sublimation of ice London
forcesAmmoniavaporisation having hydrogen bonding
Solution
The stronger the attractions betweenparticles (molecules or ions), the more difficult
it will be toseparate the particles. When substances melt, the particles arestill close to one
another but the forces of attraction that heldthe particles rigidly together in the solid state have
beensufficiently overcome to allow the particles to move. Whensubstances boil, the particles are
completely separated from oneanother and the attractions between molecules are
completelyovercome. The energy required to cause substances to melt and toboil, and thus
disrupt the forces of attraction, comes from theenvironment surrounding the material. If you
place a piece of icein your hand, the ice will melt more quickly than if it is placedon a cold
counter top. The energy required to melt the ice comesfrom your hand, your hand gets colder and
the ice gets warmer. for ice melting hydrogen bonding for sublimation of ice London
forcesAmmoniavaporisation having hydrogen bonding.
a. It gives benzalacetone b. acetone will underg.pdf
a. It gives benzalacetone b. acetone will undergo intermolecular aldol condensation
c. benzaldehyde undergoes claisen condensation d. addition over double bond in product takes
place
Solution
a. It gives benzalacetone b. acetone will undergo intermolecular aldol condensation
c. benzaldehyde undergoes claisen condensation d. addition over double bond in product takes
place.
what is the mean, median, mode and the midrange for 191, 167,257,1.pdf
what is the mean, median, mode and the midrange for : 191,
167,257,195,381,261,235,332,225,245
Solution
Mean = Sum of the numbers/n = 3174.6
Median = Since the no of values is 15 which is odd.. the median will be the middle value i.e. 8th
value in increasing order = 1479
Mode = Value with the highest frequency = All have the same frequency so all are modes
Midrange = ( max + min ) / 2 = (203 + 9822)/2 = 5012.5.
Disodium malonate is an ionic sodium salt that is soluble in water..pdf
Disodium malonate is an ionic sodium salt that is soluble in water.
Naphthalene is a molecular covalent organic solid that is insoluble in water.
To purify, add the mixture to water and stir to dissolve the disodium malonate and then filter.
The insoluble naphthalene residue will be trapped in the filter paper, while the pure disodium
malonate can be recovered by evaporating the filtrate.
Solution
Disodium malonate is an ionic sodium salt that is soluble in water.
Naphthalene is a molecular covalent organic solid that is insoluble in water.
To purify, add the mixture to water and stir to dissolve the disodium malonate and then filter.
The insoluble naphthalene residue will be trapped in the filter paper, while the pure disodium
malonate can be recovered by evaporating the filtrate..
They showed that GFP protein could be expressed in the organisms wit.pdf
They showed that GFP protein could be expressed in the organisms with which they can be used
as a gene expression marker and protein localisation.
They used GFP in transcriptional fusions to characterize the expression pattern of genes. They
used the transformation techniques using DNA encoding GFP.
Solution
They showed that GFP protein could be expressed in the organisms with which they can be used
as a gene expression marker and protein localisation.
They used GFP in transcriptional fusions to characterize the expression pattern of genes. They
used the transformation techniques using DNA encoding GFP..
Recommended
2. Ligamentum arteriosumThe aortic arch loops over the left pulmo.pdf
2. Ligamentum arteriosum:
The aortic arch loops over the left pulmonary artery and the bifurcation of the pulmonary trunk,
to which it remains connected by the ligamentum arteriosum, a remnant of the fetal circulation
that is obliterated a few days after birth.
3. Caecum:
The cecum is a pouch-like dead-end passage that branches inferiorly from the end of the ileum
(end part of small intestine). Fecal matter entering the large intestine from the ileum passes into
the cecum before being pushed superiorly into the ascending colon. The appendix is attached to
the inferior end of the cecum and is believed to store beneficial bacteria that help break down
undigested food.
Solution
2. Ligamentum arteriosum:
The aortic arch loops over the left pulmonary artery and the bifurcation of the pulmonary trunk,
to which it remains connected by the ligamentum arteriosum, a remnant of the fetal circulation
that is obliterated a few days after birth.
3. Caecum:
The cecum is a pouch-like dead-end passage that branches inferiorly from the end of the ileum
(end part of small intestine). Fecal matter entering the large intestine from the ileum passes into
the cecum before being pushed superiorly into the ascending colon. The appendix is attached to
the inferior end of the cecum and is believed to store beneficial bacteria that help break down
undigested food..
moles should be equal to complete consumption so.pdf
moles should be equal to complete consumption so v*0.250 = 0.150*0.100 /2 ( 2 is
for - as Pb(NO3)2 gives 2 moles of NO3- when dissociate) so v = 0.03 L ANSWER
Solution
moles should be equal to complete consumption so v*0.250 = 0.150*0.100 /2 ( 2 is
for - as Pb(NO3)2 gives 2 moles of NO3- when dissociate) so v = 0.03 L ANSWER.
TrueOne of the fundamental rules which control the PLSS system is .pdf
True
One of the fundamental rules which control the PLSS system is that the boundaries of the public
land once approved and accepted is unchangeable.
Solution
True
One of the fundamental rules which control the PLSS system is that the boundaries of the public
land once approved and accepted is unchangeable..
The stronger the attractions betweenparticles (mo.pdf
The stronger the attractions betweenparticles (molecules or ions), the more difficult
it will be toseparate the particles. When substances melt, the particles arestill close to one
another but the forces of attraction that heldthe particles rigidly together in the solid state have
beensufficiently overcome to allow the particles to move. Whensubstances boil, the particles are
completely separated from oneanother and the attractions between molecules are
completelyovercome. The energy required to cause substances to melt and toboil, and thus
disrupt the forces of attraction, comes from theenvironment surrounding the material. If you
place a piece of icein your hand, the ice will melt more quickly than if it is placedon a cold
counter top. The energy required to melt the ice comesfrom your hand, your hand gets colder and
the ice gets warmer. for ice melting hydrogen bonding for sublimation of ice London
forcesAmmoniavaporisation having hydrogen bonding
Solution
The stronger the attractions betweenparticles (molecules or ions), the more difficult
it will be toseparate the particles. When substances melt, the particles arestill close to one
another but the forces of attraction that heldthe particles rigidly together in the solid state have
beensufficiently overcome to allow the particles to move. Whensubstances boil, the particles are
completely separated from oneanother and the attractions between molecules are
completelyovercome. The energy required to cause substances to melt and toboil, and thus
disrupt the forces of attraction, comes from theenvironment surrounding the material. If you
place a piece of icein your hand, the ice will melt more quickly than if it is placedon a cold
counter top. The energy required to melt the ice comesfrom your hand, your hand gets colder and
the ice gets warmer. for ice melting hydrogen bonding for sublimation of ice London
forcesAmmoniavaporisation having hydrogen bonding.
a. It gives benzalacetone b. acetone will underg.pdf
a. It gives benzalacetone b. acetone will undergo intermolecular aldol condensation
c. benzaldehyde undergoes claisen condensation d. addition over double bond in product takes
place
Solution
a. It gives benzalacetone b. acetone will undergo intermolecular aldol condensation
c. benzaldehyde undergoes claisen condensation d. addition over double bond in product takes
place.
what is the mean, median, mode and the midrange for 191, 167,257,1.pdf
what is the mean, median, mode and the midrange for : 191,
167,257,195,381,261,235,332,225,245
Solution
Mean = Sum of the numbers/n = 3174.6
Median = Since the no of values is 15 which is odd.. the median will be the middle value i.e. 8th
value in increasing order = 1479
Mode = Value with the highest frequency = All have the same frequency so all are modes
Midrange = ( max + min ) / 2 = (203 + 9822)/2 = 5012.5.
Disodium malonate is an ionic sodium salt that is soluble in water..pdf
Disodium malonate is an ionic sodium salt that is soluble in water.
Naphthalene is a molecular covalent organic solid that is insoluble in water.
To purify, add the mixture to water and stir to dissolve the disodium malonate and then filter.
The insoluble naphthalene residue will be trapped in the filter paper, while the pure disodium
malonate can be recovered by evaporating the filtrate.
Solution
Disodium malonate is an ionic sodium salt that is soluble in water.
Naphthalene is a molecular covalent organic solid that is insoluble in water.
To purify, add the mixture to water and stir to dissolve the disodium malonate and then filter.
The insoluble naphthalene residue will be trapped in the filter paper, while the pure disodium
malonate can be recovered by evaporating the filtrate..
They showed that GFP protein could be expressed in the organisms wit.pdf
They showed that GFP protein could be expressed in the organisms with which they can be used
as a gene expression marker and protein localisation.
They used GFP in transcriptional fusions to characterize the expression pattern of genes. They
used the transformation techniques using DNA encoding GFP.
Solution
They showed that GFP protein could be expressed in the organisms with which they can be used
as a gene expression marker and protein localisation.
They used GFP in transcriptional fusions to characterize the expression pattern of genes. They
used the transformation techniques using DNA encoding GFP..
Thepossible mechanism by which a 75kD nuclear protein is imported in.pdf
Thepossible mechanism by which a 75kD nuclear protein is imported into teh nucleus includes.
In eukaryotic cells the nucleus is physically separated from the cytoplasm by a double membrane
structure, the nuclear envelope (NE). The NE is crossed by multiple supramolecular structures
specialized in mediating the bidirectional traffic of molecules between the nucleus and the
cytoplasm. These structures are designated the nuclear pore complex (NPC), (Freitas, 2009).
Thus the nuclear pore complex (NPC), perhaps the largest protein complex in the cell, is
responsible for the protected exchange of components between the nucleus and cytoplasm and
for preventing the transport of material not destined to cross the nuclear envelope(UIUC).
Transport Cycle.In order to pass through the NPC, a large molecule (cargo) i.e. theteh cargo
proteinmust associate with another protein called a transport receptor. These transport receptors
then act as chaperones which shuttle cargo through the NPC. Recognition by the transport
receptor takes place via a specific sequence of amino acids in the cargo protein. Besides cargo
and transport receptors, the third ingredient necessary for nuclear transport is the signaling
protein Ran, which hydrolyzes GTP. Ran is responsible for regulating the interaction of transport
receptor and cargo and RanGDP/RanGTP concentration gradients across the nuclear envelope
drive nuclear import and export. Once a transport receptor identifies and binds its cargo, the
transport complex may be imported or exported through the NPC. Repeating amino acid
sequences involving phenyalanine (F) and glycine (G) have been thoroughly implicated in the
process selective import,(UIUC).
Solution
Thepossible mechanism by which a 75kD nuclear protein is imported into teh nucleus includes.
In eukaryotic cells the nucleus is physically separated from the cytoplasm by a double membrane
structure, the nuclear envelope (NE). The NE is crossed by multiple supramolecular structures
specialized in mediating the bidirectional traffic of molecules between the nucleus and the
cytoplasm. These structures are designated the nuclear pore complex (NPC), (Freitas, 2009).
Thus the nuclear pore complex (NPC), perhaps the largest protein complex in the cell, is
responsible for the protected exchange of components between the nucleus and cytoplasm and
for preventing the transport of material not destined to cross the nuclear envelope(UIUC).
Transport Cycle.In order to pass through the NPC, a large molecule (cargo) i.e. theteh cargo
proteinmust associate with another protein called a transport receptor. These transport receptors
then act as chaperones which shuttle cargo through the NPC. Recognition by the transport
receptor takes place via a specific sequence of amino acids in the cargo protein. Besides cargo
and transport receptors, the third ingredient necessary for nuclear transport is the signaling
protein Ran, which hydrolyzes GTP. Ran is responsible for regulating the inter.
The neurons are nervous cells which play the central role in generat.pdf
The neurons are nervous cells which play the central role in generating, holding and transmitting
information across the body. The information as we perceive is a physico-physiological aspect of
our reaction to an action but it is comprised of a complex set of cytological, electro-chemical and
bio-molecular state of entitites. To be specific, the mode of action of neurons can be functionally
understood as:
The neurons comprise of a complex set of neurological membranes and ion-channels. When a
stimulus is experienced by the body/brain, it changes the bio-chemical nature/concentrations of
various electrically stimulated ions present inside the neurons and outside their perphery. This
causes generation of a local ionic gradient in the neurons which further drives physical activity
of the neuron. The neurological membranes play important role in relay of information. These
membranes help in relay of these biochemical signals across various neurological membranes
and thus transit this information from one neuron to another via a matrix between the two, called
synapse. This process takes place in a very small fraction of seconds so that an immediate and
appropriate reaction could be made.
Thus, a neurological cell holds the information by means of generation of a local bio-chemical
potential and ionic gradient and relays/moves this information by means of channelizing these
local gradients in the form of electro-chemical signals from one neuron to the other through
synapse.
Solution
The neurons are nervous cells which play the central role in generating, holding and transmitting
information across the body. The information as we perceive is a physico-physiological aspect of
our reaction to an action but it is comprised of a complex set of cytological, electro-chemical and
bio-molecular state of entitites. To be specific, the mode of action of neurons can be functionally
understood as:
The neurons comprise of a complex set of neurological membranes and ion-channels. When a
stimulus is experienced by the body/brain, it changes the bio-chemical nature/concentrations of
various electrically stimulated ions present inside the neurons and outside their perphery. This
causes generation of a local ionic gradient in the neurons which further drives physical activity
of the neuron. The neurological membranes play important role in relay of information. These
membranes help in relay of these biochemical signals across various neurological membranes
and thus transit this information from one neuron to another via a matrix between the two, called
synapse. This process takes place in a very small fraction of seconds so that an immediate and
appropriate reaction could be made.
Thus, a neurological cell holds the information by means of generation of a local bio-chemical
potential and ionic gradient and relays/moves this information by means of channelizing these
local gradients in the form of electro-chemical signals from one neuron to the other through
synapse..
The parts of the respiratory tract which have nonkeratinized stratif.pdf
The parts of the respiratory tract which have nonkeratinized stratified squamous epithelium are
Larynx-vocal cords, Oropharynx and Laryngopharynx.
Solution
The parts of the respiratory tract which have nonkeratinized stratified squamous epithelium are
Larynx-vocal cords, Oropharynx and Laryngopharynx..
The impact of Brexit would be visible on the financial instituations.pdf
The impact of Brexit would be visible on the financial instituations of US and Europe. The
impact would be shorter and longer term both. The biggest of all is the uncertainty it has created.
So at first the financial firms would hold on to the likely investments they are going to make in
UK as they would go in wait and watch mode.
Also they are likely going to wrap up the operations or shift their base from UK to other
European country assuming that the base operations caters primarily to the European nations.
This also leads to a number of Mergers and acquisitions as the valuations of a lot of businesses
would turn attractive due to currency depreciation.
Solution
The impact of Brexit would be visible on the financial instituations of US and Europe. The
impact would be shorter and longer term both. The biggest of all is the uncertainty it has created.
So at first the financial firms would hold on to the likely investments they are going to make in
UK as they would go in wait and watch mode.
Also they are likely going to wrap up the operations or shift their base from UK to other
European country assuming that the base operations caters primarily to the European nations.
This also leads to a number of Mergers and acquisitions as the valuations of a lot of businesses
would turn attractive due to currency depreciation..
The alkaline earth metals burn in oxygen to form the monoxide, MO wh.pdf
The alkaline earth metals burn in oxygen to form the monoxide, MO which, except for BeO,
have rock-salt structure (BeO has wurtzite structure). The BeO is essentially covalent in nature.
The enthalpies of formation of these oxides are quite high and consequently they are very stable
to heat. BeO is amphoteric while oxides of other elements are ionic in nature. All these oxides
except BeO are basic in nature and react with water to form sparingly soluble hydroxides.
Solution
The alkaline earth metals burn in oxygen to form the monoxide, MO which, except for BeO,
have rock-salt structure (BeO has wurtzite structure). The BeO is essentially covalent in nature.
The enthalpies of formation of these oxides are quite high and consequently they are very stable
to heat. BeO is amphoteric while oxides of other elements are ionic in nature. All these oxides
except BeO are basic in nature and react with water to form sparingly soluble hydroxides..
Squid molluscCrab ArthropodsCrayfish ArthropodsPolychaete .pdf
Squid: mollusc
Crab: Arthropods
Crayfish: Arthropods
Polychaete: Annelids
Homologous organs are the one that develop from common ancestor and have diverged to
perform different functions according to the habitat and habit of the organisms. Lateral
appendages present in Crab (arthropod), crayfish (arthropod) and polychaete (annelid) exhibit
homology. The lateral appendages (parapodia) in polychaete worms extend from the body
segments and serve in locomotion. However, soft body plan does not fix them firmly. On the
other hand, arthropods have hard bodied skeleton. Crab and crayfish takes advantage of hard
bodied skeleton and their lateral appendages are fixed firmly in the skeleton to provide maximum
leverage. Development of parapodia in polychetes allowed them to thrive out of burrow in the
muddy oceanic sediments.
Solution
Squid: mollusc
Crab: Arthropods
Crayfish: Arthropods
Polychaete: Annelids
Homologous organs are the one that develop from common ancestor and have diverged to
perform different functions according to the habitat and habit of the organisms. Lateral
appendages present in Crab (arthropod), crayfish (arthropod) and polychaete (annelid) exhibit
homology. The lateral appendages (parapodia) in polychaete worms extend from the body
segments and serve in locomotion. However, soft body plan does not fix them firmly. On the
other hand, arthropods have hard bodied skeleton. Crab and crayfish takes advantage of hard
bodied skeleton and their lateral appendages are fixed firmly in the skeleton to provide maximum
leverage. Development of parapodia in polychetes allowed them to thrive out of burrow in the
muddy oceanic sediments..
The code you written is little bit confusing and lengthy. Here is th.pdf
The code you written is little bit confusing and lengthy. Here is the efficient code to your
problem.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class ZooMonitor {
private static Scanner scanner=new Scanner(System.in);
public static void main(String[] args) {
int choice;
while(true){
System.out.println(\"------ Menu ------\");
System.out.println(\"1. Monitor an animal\");
System.out.println(\"2. Monitor a habitat\");
System.out.println(\"3. Exit\");
choice=scanner.nextInt();
switch(choice){
case 1:
monitorAnimal();
break;
case 2:
monitorHabitat();
break;
case 3:
System.exit(0);
default:
System.out.println(\"Invalid Input\");
}
}
}
public static void monitorAnimal(){
int choice;
BufferedReader br = null;
int count=0;
Map animals=new HashMap();
System.out.println(\"------ Animals ------\");
try {
String sCurrentLine;
br = new BufferedReader(new FileReader(\"animals.txt\"));
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Details\")){
String animal=sCurrentLine.split(\" \")[2];
System.out.println(++count+\". \"+animal);
animals.put(count, animal);
}
}
br.close();
br = new BufferedReader(new FileReader(\"animals.txt\"));
System.out.println(\"Enter choice : \");
choice=scanner.nextInt();
String selectedAnimal=animals.get(choice);
if(selectedAnimal!=null){
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Animal\")&&(sCurrentLine.split(\"
\")[2].equalsIgnoreCase(selectedAnimal)||selectedAnimal.startsWith(sCurrentLine.split(\"
\")[2].toLowerCase()))){
for(int i=0;i<5;i++){
if(sCurrentLine.startsWith(\"*\"))
System.err.println(sCurrentLine.substring(5));
else
System.out.println(sCurrentLine);
sCurrentLine=br.readLine();
}
}
}
}
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
if (br != null)br.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
}
public static void monitorHabitat(){
int choice;
BufferedReader br = null;
int count=0;
Map habitats=new HashMap();
System.out.println(\"------ Habitats ------\");
try {
String sCurrentLine;
br = new BufferedReader(new FileReader(\"habitats.txt\"));
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Details\")){
String habitat=sCurrentLine.split(\" \")[2];
System.out.println(++count+\". \"+habitat);
habitats.put(count, habitat);
}
}
br.close();
br = new BufferedReader(new FileReader(\"habitats.txt\"));
System.out.println(\"Enter choice : \");
choice=scanner.nextInt();
String selectedAnimal=habitats.get(choice);
if(selectedAnimal!=null){
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Habitat\")&&(sCurrentLine.split(\"
\")[2].equalsIgnoreCase(selectedAnimal)||selectedAnimal.startsWith(sCurrentLine.split(\"
\")[2].toLowerCase()))){
for(int i=0;i<4;i++){
if(sCurrentLine.startsWith(\"*\"))
System.err.println(sCurrentLine.substring(5));
else
System.out..
Ques-1Normala flora and microbiota of nose and mouth are non-infe.pdf
Ques-1:
Normala flora and microbiota of nose and mouth are non-infectious in the region where they are
living in colonies at specific pH and temperature. However, if they reach other parts of human
body other than nose and mouth may induce “infection”
They do possess common characteristics of sensing known as “quorum sensing” in which
bacteria flora are going to communicate with each other in the region or habitat where they are
living. They have a specific property of “horizontal gene transfer”
Staphylococcus epidermis & lactobacillus are going to present in mouth Corynebacterium and
spirochetes are going to present in nose due to presence of moist environments
Ques-2:
Answer:
Yes, there are other microorganisms inside the mouth & nose other normal flora and microbiota.
There are some fungal microbial pathogens, spirochetes, aspergillus, blastomycosis at the time of
infection present in oral & nasal cavities
Solution
Ques-1:
Normala flora and microbiota of nose and mouth are non-infectious in the region where they are
living in colonies at specific pH and temperature. However, if they reach other parts of human
body other than nose and mouth may induce “infection”
They do possess common characteristics of sensing known as “quorum sensing” in which
bacteria flora are going to communicate with each other in the region or habitat where they are
living. They have a specific property of “horizontal gene transfer”
Staphylococcus epidermis & lactobacillus are going to present in mouth Corynebacterium and
spirochetes are going to present in nose due to presence of moist environments
Ques-2:
Answer:
Yes, there are other microorganisms inside the mouth & nose other normal flora and microbiota.
There are some fungal microbial pathogens, spirochetes, aspergillus, blastomycosis at the time of
infection present in oral & nasal cavities.
Solution.According to IASB framework INCOME is the one of element .pdf
Solution
.
According to IASB framework INCOME is the one of element of balance sheet which
increases in economic benefits during the accounting period in the form of inflows or
enhancements of assets or decreases of liabilities that result in increases in equity, other than
those relating to contributions from equity participants?.
7. Answer is A.Cytochromes are protien containing iron or heme gro.pdf
7. Answer is A.
Cytochromes are protien containing iron or heme groups and function as electron transfer agents
in many metabolic pathways in mitochondria and chloroplasts. It doesn\'t pump ATP, It doesn\'t
has sulfur groups and it doesn\'t reduce oxygen to water.
8. Answer is B.
Pyruvate to is converted to acetyl-coA by pyruvate dehydrogenase complex. First Pyruvate is
decarboxylated by pyruvate dehydrogenase with help from TPP. Then dihydrolipoyl
transacetylase with cofactor lipoamide oxidizes hydroxyethyl- to acetyl- and then transfers
acetyl- to CoA, forming acetyl-CoA. Finally dihydrolipoyl dehydrogenase oxidizes the thiol
groups of the dihydrolipoamide back to lipoamide reducing NAD to NADH.
9. Answer is E.
The strongest oxidizing agent is oxygen, the oxidation of reduced NADH by oxygen has
potential difference of 1.136 V; the electron flow will be from the most negative system, which
is the couple NAD+/NADH, to the least negative system which is the couple O2/H2O. Thus the
oxidation of the reduced form, NADH,oxygen is a spontaneous and a slow reaction, unless
enzymes are present to catalyze the reaction.
Solution
7. Answer is A.
Cytochromes are protien containing iron or heme groups and function as electron transfer agents
in many metabolic pathways in mitochondria and chloroplasts. It doesn\'t pump ATP, It doesn\'t
has sulfur groups and it doesn\'t reduce oxygen to water.
8. Answer is B.
Pyruvate to is converted to acetyl-coA by pyruvate dehydrogenase complex. First Pyruvate is
decarboxylated by pyruvate dehydrogenase with help from TPP. Then dihydrolipoyl
transacetylase with cofactor lipoamide oxidizes hydroxyethyl- to acetyl- and then transfers
acetyl- to CoA, forming acetyl-CoA. Finally dihydrolipoyl dehydrogenase oxidizes the thiol
groups of the dihydrolipoamide back to lipoamide reducing NAD to NADH.
9. Answer is E.
The strongest oxidizing agent is oxygen, the oxidation of reduced NADH by oxygen has
potential difference of 1.136 V; the electron flow will be from the most negative system, which
is the couple NAD+/NADH, to the least negative system which is the couple O2/H2O. Thus the
oxidation of the reduced form, NADH,oxygen is a spontaneous and a slow reaction, unless
enzymes are present to catalyze the reaction..
A and C. RGB analog component video uses no compression, imposes no .pdf
A and C. RGB analog component video uses no compression, imposes no real limit on color
depth and resolution, requires a large amount of bandwidth to carry a signal, and is being
superseded by DVI and HDMI, which provide better clarity.
Solution
A and C. RGB analog component video uses no compression, imposes no real limit on color
depth and resolution, requires a large amount of bandwidth to carry a signal, and is being
superseded by DVI and HDMI, which provide better clarity..
Public key authentication is the most secure colution and utilizes a.pdf
Public key authentication is the most secure colution and utilizes a public key cryptography(PKI)
system, using both public and private keys.A PKI is certainly the easiest solution from the users
prespective.It mainly supports the distribution and identification of public encryption keys,
enabling users and computers to both securely exchange data over networks such as the internet
and verify the identity of the other party.
Without PKI , sensitive information can still be encrypted ( ensuring confidentiality) and
exchanged but there would be no assurance of the identity(authentication) of the other party. Any
form of sensitive data exchanged over the internet is relianton PKI for security.
Elements of PKI
A typical PKI consists of hardware, software , policies and standards to manage the
creation,admisinstration , distribution and revocation of keys and digital certificates. Digital
certificates are at the heart of PKI as they affrim the identity of the certificate subject that bind
the identity to the public key contained in the certificate.
Following are the elements of the PKI:
Certificate Authority (CA)
Registration authority
A certificate database , it mainly stores the certificate requests and issues and revokes
certificates.
A certificate store , which resides on a local computer as a place to store issues certificates and
private keys.
--------------------------************************-----------------------------
To connect many distant employees at once, all office locations must be able to access the same
network resource. Most time iof we will would house the core infrastucture needed to establish
the network, such as the servers and databases. Generally Wide area network (WAN) to connect
the remote offices. WAN are used to connect Local area network(LAN) for disparate offices
together.
However installation and overhead of LNS and WANS is expensive.
To over come this Virtual private networks that run over the internet.
VPN it mainly provides secure connections between individual users and their organizations
network over and their organizations network over the internet, have several upsides. They
provide more secure site to site connections. They provide information transfer much faster that
WAN\'s and most importantly to small and medimum sized business, VPN\'s are much
lessexpensive since you can use a single leased line to the internet for each ofice, cutting down
on broadband cost.
While VPN\'s are highly recommended and trusted a new technology gaining traction with
compaines is the remote desktop. Remote dsktops requires software or an operating system that
allows aplications to run remotely on a server, but to be displayed locally simultaneously. This
network server hosts remote files and application in a secure location , ensuring your data is
never lost , even when something happens to device you are working on.
Solution
Public key authentication is the most secure colution and utilizes a public key cryptography(PKI)
sy.
Particle Charging Mechanisms in ESPParticles are charged by negat.pdf
Particle Charging Mechanisms in ESP:
Particles are charged by negative gas ions moving toward the collection plate by one of these two
mechanisms:
1. Field charging 2. diffusion charging.
In field charging, particles capture negatively charged gas ions as the ions move toward the
grounded collection plate. Diffusion charging, as its name implies, depends on the random
motion of the gas ions to charge particles. In field charging, as particles enter the electric field,
they cause a local dislocation of the field. Negative gas ions traveling along the electric field
lines collide with the suspended particles and impart a charge to them. The ions will continue to
bombard a particle until the charge on that particle is sufficient to divert the electric lines away
from it. This prevents new ions from colliding with the charged dust particle. When a particle no
longer receives an ion charge, it is said to be saturated. Saturated charged particles then migrate
to the collection electrode and are collected.
Diffusion charging is associated with the random Brownian motion of the negative gas ions. The
random motion is related to the velocity of the gas ions due to thermal effects: the higher the
temperature, the more movement. Negative gas ions collide with the particles because of their
random thermal motion and impart a charge on the particles. Because the particles are very
small, they do not cause the electric field to be dislocated, as in field charging. Thus, diffusion
charging is the only mechanism by which these very small particles become charged. The
charged particles then migrate to the collection electrode. Each of these two charging
mechanisms occurs to some extent, with one dominating depending on particle size.
Field charging dominates for particles with a diameter >1.0 micrometer because particles must
be large enough to capture gas ions. Diffusion charging dominates for particles with a diameter
less than 0.1 micrometer. A combination of these two charging mechanisms occurs for particles
ranging between 0.2 and 1.0 micrometer in diameter.
Solution
Particle Charging Mechanisms in ESP:
Particles are charged by negative gas ions moving toward the collection plate by one of these two
mechanisms:
1. Field charging 2. diffusion charging.
In field charging, particles capture negatively charged gas ions as the ions move toward the
grounded collection plate. Diffusion charging, as its name implies, depends on the random
motion of the gas ions to charge particles. In field charging, as particles enter the electric field,
they cause a local dislocation of the field. Negative gas ions traveling along the electric field
lines collide with the suspended particles and impart a charge to them. The ions will continue to
bombard a particle until the charge on that particle is sufficient to divert the electric lines away
from it. This prevents new ions from colliding with the charged dust particle. When a particle no
longer receives an ion charge, it is.
Please follow the data There are some flag and the control charac.pdf
Please follow the data :
There are some flag and the control characters that are usewd in the fields of communication to
maintain a synchronous data from the sender and the receiver. Some of the flags described are
the ACK and NAK flags or the control characters. These characters will generally have the
standard decimal values that maintain.
a) ACK : It is the flag value that assigns to acknowledge indicating the result of the process that
has been initiated from the user or the sender. It also describes that the process was successful. It
is the abbrevation of Acknowledgement.
The decimal value for the ACK flag is given by (6)10. So from the conversion of decimal to
binary we get the values of the ACK flag as (0110)2.
Now the hexadecimal value can be given as the same as the values for the hexa and the binary
would be same till the value of 15 as though they describr the same point. So the hexadecimal
value of the ACK flag is (6)16.
b) NAK : It is a flag or a boolean value that describes the sender or the user indicating that the
data has not been received correctly or the data has been tampered or lost over the network. It is
the abbrevation of Negative Acknowledgement.
The decimal value that is listed accordingly for the characters is given by (21)10. So the binary
format of the NAK flag is givem by (0001 0101)2.
Now that converting it to the hexadecimal value we get the NAK flag value as 15(16).
Hope this is helpful.
Solution
Please follow the data :
There are some flag and the control characters that are usewd in the fields of communication to
maintain a synchronous data from the sender and the receiver. Some of the flags described are
the ACK and NAK flags or the control characters. These characters will generally have the
standard decimal values that maintain.
a) ACK : It is the flag value that assigns to acknowledge indicating the result of the process that
has been initiated from the user or the sender. It also describes that the process was successful. It
is the abbrevation of Acknowledgement.
The decimal value for the ACK flag is given by (6)10. So from the conversion of decimal to
binary we get the values of the ACK flag as (0110)2.
Now the hexadecimal value can be given as the same as the values for the hexa and the binary
would be same till the value of 15 as though they describr the same point. So the hexadecimal
value of the ACK flag is (6)16.
b) NAK : It is a flag or a boolean value that describes the sender or the user indicating that the
data has not been received correctly or the data has been tampered or lost over the network. It is
the abbrevation of Negative Acknowledgement.
The decimal value that is listed accordingly for the characters is given by (21)10. So the binary
format of the NAK flag is givem by (0001 0101)2.
Now that converting it to the hexadecimal value we get the NAK flag value as 15(16).
Hope this is helpful..
only anhydorus LiAlH4 acts as a good reducing agent ,that is i.pdf
only anhydorus LiAlH4 acts as a good reducing agent ,
that is in the absence of H20 molecules ,
when water is present , its reducing proertiese are lessened <------ans
Solution
only anhydorus LiAlH4 acts as a good reducing agent ,
that is in the absence of H20 molecules ,
when water is present , its reducing proertiese are lessened <------ans.
Part 1)#include stdio.h #include stdlib.h #include pthrea.pdf
Part 1)
#include
#include
#include
#include
/*Error handling for pthread_create and pthread_join*/
/*from the pthread_create man page*/
#define handle_error_en(en, msg) \\
do { errno = en; perror(msg); exit(EXIT_FAILURE); } while (0)
/* # of running threads */
volatile int running_threads = 0;
pthread_t thread[3]; /*Descriptors for our 3 threads*/
int numOfElements;/*Total # of elements from the user*/
struct Results{ /*Struct to hold the statistical results*/
int min;
int max;
int average;
}Results;
/*This function finds the minimum element of an array*/
void *findMin(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
Results.min = elements[0]; /*set minimum to first element */
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] < Results.min){ /*if the current element is less than the current min*/
Results.min = elements[i]; /*store the new min*/
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the maximum element of an array*/
void *findMax(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] > Results.max){ /*store the new max*/
Results.max = elements[i];
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the average of an array*/
void *findAverage(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
Results.average += elements[i]; /*add element @ i to average*/
}
Results.average = Results.average/numOfElements; /*Divide the sum by the number of
elements*/
running_threads -= 1; /*Decrement running threads counter*/
return NULL;
}
/* This method accepts a int n(initial size of array) and
pointer to an array and returns # of elements in the array
*/
int getArrayInput(int n, int *array_ptr){
int input;/*Store user input */
int numberOfElements = 0;/*Number of Integers inputed*/
printf(\"Creating Dynamic Array...\ -\ \");
for(;;){ /*infinite loop*/
printf(\"Enter a positive value:\ Negative Number to Stop\ -\ \");
//Get Int from console, store at address of input
if (scanf(\"%d\",&input) != 1){
printf(\"\ Oops that wasn\'t an Integer\ lets try filling the array again\ Remember
INTEGERS only!\ \");
exit(EXIT_FAILURE);
}
if (input >= 0){
if (numberOfElements == n){
n += 1; //Make room for the current input
array_ptr = realloc(array_ptr, n * sizeof(int));//realloc array and set pointer
}
array_ptr[numberOfElements++] = input;//Store input at next empty element
} else {
printf(\"\ Number of Integers: %d\ \", numberOfElements);
break;
}
}
return numberOfElements;
}
/*This function joins our n number of threads */
void joinThreads(int numberOfThreads){
int i; /*count*/
int s; /*error #*/
while(numberOfThreads >= 0){ /*Join our threads*/
s = pthread_jo.
Note that there are two poles in the system. The common region of th.pdf
Note that there are two poles in the system. The common region of the entire system is inside the
one of the poles. Therefore, the region of convergence is less than the unit circle. Therefore, the
signal is left-sided.
If it is greater than unit circuit, it is a right-sided signal.
If it is between the poles, it is two-sided.
Solution
Note that there are two poles in the system. The common region of the entire system is inside the
one of the poles. Therefore, the region of convergence is less than the unit circle. Therefore, the
signal is left-sided.
If it is greater than unit circuit, it is a right-sided signal.
If it is between the poles, it is two-sided..
Obvious first to third shell. E is proportional .pdf
Obvious first to third shell. E is proportional to [1/n1^2 -1/n2^2] [1/n1^2 -1/n2^2]
is more when electron moves from the first to the third shel
Solution
Obvious first to third shell. E is proportional to [1/n1^2 -1/n2^2] [1/n1^2 -1/n2^2]
is more when electron moves from the first to the third shel.
a) YES. Both are hydrocarbons and nonpolar. Like .pdf
a) YES. Both are hydrocarbons and nonpolar. Like dissolves like. b) NO. CBr4 is
nonpolar and H2O is polar. Both do not mix. c) YES. LiNO3 is ionic and H2O is a polar solvent.
Polar solvents dissolve polar and ionic solutes. d) No. CH3OH is polar due to the O-H bond and
the other molecule is a hydrocarbon and nonpolar. They do not mix this would be helpful
Solution
a) YES. Both are hydrocarbons and nonpolar. Like dissolves like. b) NO. CBr4 is
nonpolar and H2O is polar. Both do not mix. c) YES. LiNO3 is ionic and H2O is a polar solvent.
Polar solvents dissolve polar and ionic solutes. d) No. CH3OH is polar due to the O-H bond and
the other molecule is a hydrocarbon and nonpolar. They do not mix this would be helpful.
Basic assumptions 1 Pairs of electrons in the va.pdf
Basic assumptions 1 Pairs of electrons in the valence shell of a central atom repel
each other. 2 These pairs of electrons tend to occupy positions in space that minimize repulsions
and maximize the distance of separation between them. 3 The valence shell is taken as a sphere
with electron pairs localizing on the spherical surface at maximum distance from one another. 4
A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a
multiple bond are treated as a single super pair. 5 Where two or more resonance structures can
depict a molecule the VSEPR model is applicable to any such structure.
Solution
Basic assumptions 1 Pairs of electrons in the valence shell of a central atom repel
each other. 2 These pairs of electrons tend to occupy positions in space that minimize repulsions
and maximize the distance of separation between them. 3 The valence shell is taken as a sphere
with electron pairs localizing on the spherical surface at maximum distance from one another. 4
A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a
multiple bond are treated as a single super pair. 5 Where two or more resonance structures can
depict a molecule the VSEPR model is applicable to any such structure..
The geography of Taylor Swift - some ideas
Geographical themes connected with Taylor Swift's ERAS tour - coming to the UK in June 2024
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Thepossible mechanism by which a 75kD nuclear protein is imported in.pdf
Thepossible mechanism by which a 75kD nuclear protein is imported into teh nucleus includes.
In eukaryotic cells the nucleus is physically separated from the cytoplasm by a double membrane
structure, the nuclear envelope (NE). The NE is crossed by multiple supramolecular structures
specialized in mediating the bidirectional traffic of molecules between the nucleus and the
cytoplasm. These structures are designated the nuclear pore complex (NPC), (Freitas, 2009).
Thus the nuclear pore complex (NPC), perhaps the largest protein complex in the cell, is
responsible for the protected exchange of components between the nucleus and cytoplasm and
for preventing the transport of material not destined to cross the nuclear envelope(UIUC).
Transport Cycle.In order to pass through the NPC, a large molecule (cargo) i.e. theteh cargo
proteinmust associate with another protein called a transport receptor. These transport receptors
then act as chaperones which shuttle cargo through the NPC. Recognition by the transport
receptor takes place via a specific sequence of amino acids in the cargo protein. Besides cargo
and transport receptors, the third ingredient necessary for nuclear transport is the signaling
protein Ran, which hydrolyzes GTP. Ran is responsible for regulating the interaction of transport
receptor and cargo and RanGDP/RanGTP concentration gradients across the nuclear envelope
drive nuclear import and export. Once a transport receptor identifies and binds its cargo, the
transport complex may be imported or exported through the NPC. Repeating amino acid
sequences involving phenyalanine (F) and glycine (G) have been thoroughly implicated in the
process selective import,(UIUC).
Solution
Thepossible mechanism by which a 75kD nuclear protein is imported into teh nucleus includes.
In eukaryotic cells the nucleus is physically separated from the cytoplasm by a double membrane
structure, the nuclear envelope (NE). The NE is crossed by multiple supramolecular structures
specialized in mediating the bidirectional traffic of molecules between the nucleus and the
cytoplasm. These structures are designated the nuclear pore complex (NPC), (Freitas, 2009).
Thus the nuclear pore complex (NPC), perhaps the largest protein complex in the cell, is
responsible for the protected exchange of components between the nucleus and cytoplasm and
for preventing the transport of material not destined to cross the nuclear envelope(UIUC).
Transport Cycle.In order to pass through the NPC, a large molecule (cargo) i.e. theteh cargo
proteinmust associate with another protein called a transport receptor. These transport receptors
then act as chaperones which shuttle cargo through the NPC. Recognition by the transport
receptor takes place via a specific sequence of amino acids in the cargo protein. Besides cargo
and transport receptors, the third ingredient necessary for nuclear transport is the signaling
protein Ran, which hydrolyzes GTP. Ran is responsible for regulating the inter.
The neurons are nervous cells which play the central role in generat.pdf
The neurons are nervous cells which play the central role in generating, holding and transmitting
information across the body. The information as we perceive is a physico-physiological aspect of
our reaction to an action but it is comprised of a complex set of cytological, electro-chemical and
bio-molecular state of entitites. To be specific, the mode of action of neurons can be functionally
understood as:
The neurons comprise of a complex set of neurological membranes and ion-channels. When a
stimulus is experienced by the body/brain, it changes the bio-chemical nature/concentrations of
various electrically stimulated ions present inside the neurons and outside their perphery. This
causes generation of a local ionic gradient in the neurons which further drives physical activity
of the neuron. The neurological membranes play important role in relay of information. These
membranes help in relay of these biochemical signals across various neurological membranes
and thus transit this information from one neuron to another via a matrix between the two, called
synapse. This process takes place in a very small fraction of seconds so that an immediate and
appropriate reaction could be made.
Thus, a neurological cell holds the information by means of generation of a local bio-chemical
potential and ionic gradient and relays/moves this information by means of channelizing these
local gradients in the form of electro-chemical signals from one neuron to the other through
synapse.
Solution
The neurons are nervous cells which play the central role in generating, holding and transmitting
information across the body. The information as we perceive is a physico-physiological aspect of
our reaction to an action but it is comprised of a complex set of cytological, electro-chemical and
bio-molecular state of entitites. To be specific, the mode of action of neurons can be functionally
understood as:
The neurons comprise of a complex set of neurological membranes and ion-channels. When a
stimulus is experienced by the body/brain, it changes the bio-chemical nature/concentrations of
various electrically stimulated ions present inside the neurons and outside their perphery. This
causes generation of a local ionic gradient in the neurons which further drives physical activity
of the neuron. The neurological membranes play important role in relay of information. These
membranes help in relay of these biochemical signals across various neurological membranes
and thus transit this information from one neuron to another via a matrix between the two, called
synapse. This process takes place in a very small fraction of seconds so that an immediate and
appropriate reaction could be made.
Thus, a neurological cell holds the information by means of generation of a local bio-chemical
potential and ionic gradient and relays/moves this information by means of channelizing these
local gradients in the form of electro-chemical signals from one neuron to the other through
synapse..
The parts of the respiratory tract which have nonkeratinized stratif.pdf
The parts of the respiratory tract which have nonkeratinized stratified squamous epithelium are
Larynx-vocal cords, Oropharynx and Laryngopharynx.
Solution
The parts of the respiratory tract which have nonkeratinized stratified squamous epithelium are
Larynx-vocal cords, Oropharynx and Laryngopharynx..
The impact of Brexit would be visible on the financial instituations.pdf
The impact of Brexit would be visible on the financial instituations of US and Europe. The
impact would be shorter and longer term both. The biggest of all is the uncertainty it has created.
So at first the financial firms would hold on to the likely investments they are going to make in
UK as they would go in wait and watch mode.
Also they are likely going to wrap up the operations or shift their base from UK to other
European country assuming that the base operations caters primarily to the European nations.
This also leads to a number of Mergers and acquisitions as the valuations of a lot of businesses
would turn attractive due to currency depreciation.
Solution
The impact of Brexit would be visible on the financial instituations of US and Europe. The
impact would be shorter and longer term both. The biggest of all is the uncertainty it has created.
So at first the financial firms would hold on to the likely investments they are going to make in
UK as they would go in wait and watch mode.
Also they are likely going to wrap up the operations or shift their base from UK to other
European country assuming that the base operations caters primarily to the European nations.
This also leads to a number of Mergers and acquisitions as the valuations of a lot of businesses
would turn attractive due to currency depreciation..
The alkaline earth metals burn in oxygen to form the monoxide, MO wh.pdf
The alkaline earth metals burn in oxygen to form the monoxide, MO which, except for BeO,
have rock-salt structure (BeO has wurtzite structure). The BeO is essentially covalent in nature.
The enthalpies of formation of these oxides are quite high and consequently they are very stable
to heat. BeO is amphoteric while oxides of other elements are ionic in nature. All these oxides
except BeO are basic in nature and react with water to form sparingly soluble hydroxides.
Solution
The alkaline earth metals burn in oxygen to form the monoxide, MO which, except for BeO,
have rock-salt structure (BeO has wurtzite structure). The BeO is essentially covalent in nature.
The enthalpies of formation of these oxides are quite high and consequently they are very stable
to heat. BeO is amphoteric while oxides of other elements are ionic in nature. All these oxides
except BeO are basic in nature and react with water to form sparingly soluble hydroxides..
Squid molluscCrab ArthropodsCrayfish ArthropodsPolychaete .pdf
Squid: mollusc
Crab: Arthropods
Crayfish: Arthropods
Polychaete: Annelids
Homologous organs are the one that develop from common ancestor and have diverged to
perform different functions according to the habitat and habit of the organisms. Lateral
appendages present in Crab (arthropod), crayfish (arthropod) and polychaete (annelid) exhibit
homology. The lateral appendages (parapodia) in polychaete worms extend from the body
segments and serve in locomotion. However, soft body plan does not fix them firmly. On the
other hand, arthropods have hard bodied skeleton. Crab and crayfish takes advantage of hard
bodied skeleton and their lateral appendages are fixed firmly in the skeleton to provide maximum
leverage. Development of parapodia in polychetes allowed them to thrive out of burrow in the
muddy oceanic sediments.
Solution
Squid: mollusc
Crab: Arthropods
Crayfish: Arthropods
Polychaete: Annelids
Homologous organs are the one that develop from common ancestor and have diverged to
perform different functions according to the habitat and habit of the organisms. Lateral
appendages present in Crab (arthropod), crayfish (arthropod) and polychaete (annelid) exhibit
homology. The lateral appendages (parapodia) in polychaete worms extend from the body
segments and serve in locomotion. However, soft body plan does not fix them firmly. On the
other hand, arthropods have hard bodied skeleton. Crab and crayfish takes advantage of hard
bodied skeleton and their lateral appendages are fixed firmly in the skeleton to provide maximum
leverage. Development of parapodia in polychetes allowed them to thrive out of burrow in the
muddy oceanic sediments..
The code you written is little bit confusing and lengthy. Here is th.pdf
The code you written is little bit confusing and lengthy. Here is the efficient code to your
problem.
import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;
import java.util.HashMap;
import java.util.Map;
import java.util.Scanner;
public class ZooMonitor {
private static Scanner scanner=new Scanner(System.in);
public static void main(String[] args) {
int choice;
while(true){
System.out.println(\"------ Menu ------\");
System.out.println(\"1. Monitor an animal\");
System.out.println(\"2. Monitor a habitat\");
System.out.println(\"3. Exit\");
choice=scanner.nextInt();
switch(choice){
case 1:
monitorAnimal();
break;
case 2:
monitorHabitat();
break;
case 3:
System.exit(0);
default:
System.out.println(\"Invalid Input\");
}
}
}
public static void monitorAnimal(){
int choice;
BufferedReader br = null;
int count=0;
Map animals=new HashMap();
System.out.println(\"------ Animals ------\");
try {
String sCurrentLine;
br = new BufferedReader(new FileReader(\"animals.txt\"));
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Details\")){
String animal=sCurrentLine.split(\" \")[2];
System.out.println(++count+\". \"+animal);
animals.put(count, animal);
}
}
br.close();
br = new BufferedReader(new FileReader(\"animals.txt\"));
System.out.println(\"Enter choice : \");
choice=scanner.nextInt();
String selectedAnimal=animals.get(choice);
if(selectedAnimal!=null){
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Animal\")&&(sCurrentLine.split(\"
\")[2].equalsIgnoreCase(selectedAnimal)||selectedAnimal.startsWith(sCurrentLine.split(\"
\")[2].toLowerCase()))){
for(int i=0;i<5;i++){
if(sCurrentLine.startsWith(\"*\"))
System.err.println(sCurrentLine.substring(5));
else
System.out.println(sCurrentLine);
sCurrentLine=br.readLine();
}
}
}
}
} catch (IOException e) {
e.printStackTrace();
} finally {
try {
if (br != null)br.close();
} catch (IOException ex) {
ex.printStackTrace();
}
}
}
public static void monitorHabitat(){
int choice;
BufferedReader br = null;
int count=0;
Map habitats=new HashMap();
System.out.println(\"------ Habitats ------\");
try {
String sCurrentLine;
br = new BufferedReader(new FileReader(\"habitats.txt\"));
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Details\")){
String habitat=sCurrentLine.split(\" \")[2];
System.out.println(++count+\". \"+habitat);
habitats.put(count, habitat);
}
}
br.close();
br = new BufferedReader(new FileReader(\"habitats.txt\"));
System.out.println(\"Enter choice : \");
choice=scanner.nextInt();
String selectedAnimal=habitats.get(choice);
if(selectedAnimal!=null){
while ((sCurrentLine = br.readLine()) != null) {
if(sCurrentLine.startsWith(\"Habitat\")&&(sCurrentLine.split(\"
\")[2].equalsIgnoreCase(selectedAnimal)||selectedAnimal.startsWith(sCurrentLine.split(\"
\")[2].toLowerCase()))){
for(int i=0;i<4;i++){
if(sCurrentLine.startsWith(\"*\"))
System.err.println(sCurrentLine.substring(5));
else
System.out..
Ques-1Normala flora and microbiota of nose and mouth are non-infe.pdf
Ques-1:
Normala flora and microbiota of nose and mouth are non-infectious in the region where they are
living in colonies at specific pH and temperature. However, if they reach other parts of human
body other than nose and mouth may induce “infection”
They do possess common characteristics of sensing known as “quorum sensing” in which
bacteria flora are going to communicate with each other in the region or habitat where they are
living. They have a specific property of “horizontal gene transfer”
Staphylococcus epidermis & lactobacillus are going to present in mouth Corynebacterium and
spirochetes are going to present in nose due to presence of moist environments
Ques-2:
Answer:
Yes, there are other microorganisms inside the mouth & nose other normal flora and microbiota.
There are some fungal microbial pathogens, spirochetes, aspergillus, blastomycosis at the time of
infection present in oral & nasal cavities
Solution
Ques-1:
Normala flora and microbiota of nose and mouth are non-infectious in the region where they are
living in colonies at specific pH and temperature. However, if they reach other parts of human
body other than nose and mouth may induce “infection”
They do possess common characteristics of sensing known as “quorum sensing” in which
bacteria flora are going to communicate with each other in the region or habitat where they are
living. They have a specific property of “horizontal gene transfer”
Staphylococcus epidermis & lactobacillus are going to present in mouth Corynebacterium and
spirochetes are going to present in nose due to presence of moist environments
Ques-2:
Answer:
Yes, there are other microorganisms inside the mouth & nose other normal flora and microbiota.
There are some fungal microbial pathogens, spirochetes, aspergillus, blastomycosis at the time of
infection present in oral & nasal cavities.
Solution.According to IASB framework INCOME is the one of element .pdf
Solution
.
According to IASB framework INCOME is the one of element of balance sheet which
increases in economic benefits during the accounting period in the form of inflows or
enhancements of assets or decreases of liabilities that result in increases in equity, other than
those relating to contributions from equity participants?.
7. Answer is A.Cytochromes are protien containing iron or heme gro.pdf
7. Answer is A.
Cytochromes are protien containing iron or heme groups and function as electron transfer agents
in many metabolic pathways in mitochondria and chloroplasts. It doesn\'t pump ATP, It doesn\'t
has sulfur groups and it doesn\'t reduce oxygen to water.
8. Answer is B.
Pyruvate to is converted to acetyl-coA by pyruvate dehydrogenase complex. First Pyruvate is
decarboxylated by pyruvate dehydrogenase with help from TPP. Then dihydrolipoyl
transacetylase with cofactor lipoamide oxidizes hydroxyethyl- to acetyl- and then transfers
acetyl- to CoA, forming acetyl-CoA. Finally dihydrolipoyl dehydrogenase oxidizes the thiol
groups of the dihydrolipoamide back to lipoamide reducing NAD to NADH.
9. Answer is E.
The strongest oxidizing agent is oxygen, the oxidation of reduced NADH by oxygen has
potential difference of 1.136 V; the electron flow will be from the most negative system, which
is the couple NAD+/NADH, to the least negative system which is the couple O2/H2O. Thus the
oxidation of the reduced form, NADH,oxygen is a spontaneous and a slow reaction, unless
enzymes are present to catalyze the reaction.
Solution
7. Answer is A.
Cytochromes are protien containing iron or heme groups and function as electron transfer agents
in many metabolic pathways in mitochondria and chloroplasts. It doesn\'t pump ATP, It doesn\'t
has sulfur groups and it doesn\'t reduce oxygen to water.
8. Answer is B.
Pyruvate to is converted to acetyl-coA by pyruvate dehydrogenase complex. First Pyruvate is
decarboxylated by pyruvate dehydrogenase with help from TPP. Then dihydrolipoyl
transacetylase with cofactor lipoamide oxidizes hydroxyethyl- to acetyl- and then transfers
acetyl- to CoA, forming acetyl-CoA. Finally dihydrolipoyl dehydrogenase oxidizes the thiol
groups of the dihydrolipoamide back to lipoamide reducing NAD to NADH.
9. Answer is E.
The strongest oxidizing agent is oxygen, the oxidation of reduced NADH by oxygen has
potential difference of 1.136 V; the electron flow will be from the most negative system, which
is the couple NAD+/NADH, to the least negative system which is the couple O2/H2O. Thus the
oxidation of the reduced form, NADH,oxygen is a spontaneous and a slow reaction, unless
enzymes are present to catalyze the reaction..
A and C. RGB analog component video uses no compression, imposes no .pdf
A and C. RGB analog component video uses no compression, imposes no real limit on color
depth and resolution, requires a large amount of bandwidth to carry a signal, and is being
superseded by DVI and HDMI, which provide better clarity.
Solution
A and C. RGB analog component video uses no compression, imposes no real limit on color
depth and resolution, requires a large amount of bandwidth to carry a signal, and is being
superseded by DVI and HDMI, which provide better clarity..
Public key authentication is the most secure colution and utilizes a.pdf
Public key authentication is the most secure colution and utilizes a public key cryptography(PKI)
system, using both public and private keys.A PKI is certainly the easiest solution from the users
prespective.It mainly supports the distribution and identification of public encryption keys,
enabling users and computers to both securely exchange data over networks such as the internet
and verify the identity of the other party.
Without PKI , sensitive information can still be encrypted ( ensuring confidentiality) and
exchanged but there would be no assurance of the identity(authentication) of the other party. Any
form of sensitive data exchanged over the internet is relianton PKI for security.
Elements of PKI
A typical PKI consists of hardware, software , policies and standards to manage the
creation,admisinstration , distribution and revocation of keys and digital certificates. Digital
certificates are at the heart of PKI as they affrim the identity of the certificate subject that bind
the identity to the public key contained in the certificate.
Following are the elements of the PKI:
Certificate Authority (CA)
Registration authority
A certificate database , it mainly stores the certificate requests and issues and revokes
certificates.
A certificate store , which resides on a local computer as a place to store issues certificates and
private keys.
--------------------------************************-----------------------------
To connect many distant employees at once, all office locations must be able to access the same
network resource. Most time iof we will would house the core infrastucture needed to establish
the network, such as the servers and databases. Generally Wide area network (WAN) to connect
the remote offices. WAN are used to connect Local area network(LAN) for disparate offices
together.
However installation and overhead of LNS and WANS is expensive.
To over come this Virtual private networks that run over the internet.
VPN it mainly provides secure connections between individual users and their organizations
network over and their organizations network over the internet, have several upsides. They
provide more secure site to site connections. They provide information transfer much faster that
WAN\'s and most importantly to small and medimum sized business, VPN\'s are much
lessexpensive since you can use a single leased line to the internet for each ofice, cutting down
on broadband cost.
While VPN\'s are highly recommended and trusted a new technology gaining traction with
compaines is the remote desktop. Remote dsktops requires software or an operating system that
allows aplications to run remotely on a server, but to be displayed locally simultaneously. This
network server hosts remote files and application in a secure location , ensuring your data is
never lost , even when something happens to device you are working on.
Solution
Public key authentication is the most secure colution and utilizes a public key cryptography(PKI)
sy.
Particle Charging Mechanisms in ESPParticles are charged by negat.pdf
Particle Charging Mechanisms in ESP:
Particles are charged by negative gas ions moving toward the collection plate by one of these two
mechanisms:
1. Field charging 2. diffusion charging.
In field charging, particles capture negatively charged gas ions as the ions move toward the
grounded collection plate. Diffusion charging, as its name implies, depends on the random
motion of the gas ions to charge particles. In field charging, as particles enter the electric field,
they cause a local dislocation of the field. Negative gas ions traveling along the electric field
lines collide with the suspended particles and impart a charge to them. The ions will continue to
bombard a particle until the charge on that particle is sufficient to divert the electric lines away
from it. This prevents new ions from colliding with the charged dust particle. When a particle no
longer receives an ion charge, it is said to be saturated. Saturated charged particles then migrate
to the collection electrode and are collected.
Diffusion charging is associated with the random Brownian motion of the negative gas ions. The
random motion is related to the velocity of the gas ions due to thermal effects: the higher the
temperature, the more movement. Negative gas ions collide with the particles because of their
random thermal motion and impart a charge on the particles. Because the particles are very
small, they do not cause the electric field to be dislocated, as in field charging. Thus, diffusion
charging is the only mechanism by which these very small particles become charged. The
charged particles then migrate to the collection electrode. Each of these two charging
mechanisms occurs to some extent, with one dominating depending on particle size.
Field charging dominates for particles with a diameter >1.0 micrometer because particles must
be large enough to capture gas ions. Diffusion charging dominates for particles with a diameter
less than 0.1 micrometer. A combination of these two charging mechanisms occurs for particles
ranging between 0.2 and 1.0 micrometer in diameter.
Solution
Particle Charging Mechanisms in ESP:
Particles are charged by negative gas ions moving toward the collection plate by one of these two
mechanisms:
1. Field charging 2. diffusion charging.
In field charging, particles capture negatively charged gas ions as the ions move toward the
grounded collection plate. Diffusion charging, as its name implies, depends on the random
motion of the gas ions to charge particles. In field charging, as particles enter the electric field,
they cause a local dislocation of the field. Negative gas ions traveling along the electric field
lines collide with the suspended particles and impart a charge to them. The ions will continue to
bombard a particle until the charge on that particle is sufficient to divert the electric lines away
from it. This prevents new ions from colliding with the charged dust particle. When a particle no
longer receives an ion charge, it is.
Please follow the data There are some flag and the control charac.pdf
Please follow the data :
There are some flag and the control characters that are usewd in the fields of communication to
maintain a synchronous data from the sender and the receiver. Some of the flags described are
the ACK and NAK flags or the control characters. These characters will generally have the
standard decimal values that maintain.
a) ACK : It is the flag value that assigns to acknowledge indicating the result of the process that
has been initiated from the user or the sender. It also describes that the process was successful. It
is the abbrevation of Acknowledgement.
The decimal value for the ACK flag is given by (6)10. So from the conversion of decimal to
binary we get the values of the ACK flag as (0110)2.
Now the hexadecimal value can be given as the same as the values for the hexa and the binary
would be same till the value of 15 as though they describr the same point. So the hexadecimal
value of the ACK flag is (6)16.
b) NAK : It is a flag or a boolean value that describes the sender or the user indicating that the
data has not been received correctly or the data has been tampered or lost over the network. It is
the abbrevation of Negative Acknowledgement.
The decimal value that is listed accordingly for the characters is given by (21)10. So the binary
format of the NAK flag is givem by (0001 0101)2.
Now that converting it to the hexadecimal value we get the NAK flag value as 15(16).
Hope this is helpful.
Solution
Please follow the data :
There are some flag and the control characters that are usewd in the fields of communication to
maintain a synchronous data from the sender and the receiver. Some of the flags described are
the ACK and NAK flags or the control characters. These characters will generally have the
standard decimal values that maintain.
a) ACK : It is the flag value that assigns to acknowledge indicating the result of the process that
has been initiated from the user or the sender. It also describes that the process was successful. It
is the abbrevation of Acknowledgement.
The decimal value for the ACK flag is given by (6)10. So from the conversion of decimal to
binary we get the values of the ACK flag as (0110)2.
Now the hexadecimal value can be given as the same as the values for the hexa and the binary
would be same till the value of 15 as though they describr the same point. So the hexadecimal
value of the ACK flag is (6)16.
b) NAK : It is a flag or a boolean value that describes the sender or the user indicating that the
data has not been received correctly or the data has been tampered or lost over the network. It is
the abbrevation of Negative Acknowledgement.
The decimal value that is listed accordingly for the characters is given by (21)10. So the binary
format of the NAK flag is givem by (0001 0101)2.
Now that converting it to the hexadecimal value we get the NAK flag value as 15(16).
Hope this is helpful..
only anhydorus LiAlH4 acts as a good reducing agent ,that is i.pdf
only anhydorus LiAlH4 acts as a good reducing agent ,
that is in the absence of H20 molecules ,
when water is present , its reducing proertiese are lessened <------ans
Solution
only anhydorus LiAlH4 acts as a good reducing agent ,
that is in the absence of H20 molecules ,
when water is present , its reducing proertiese are lessened <------ans.
Part 1)#include stdio.h #include stdlib.h #include pthrea.pdf
Part 1)
#include
#include
#include
#include
/*Error handling for pthread_create and pthread_join*/
/*from the pthread_create man page*/
#define handle_error_en(en, msg) \\
do { errno = en; perror(msg); exit(EXIT_FAILURE); } while (0)
/* # of running threads */
volatile int running_threads = 0;
pthread_t thread[3]; /*Descriptors for our 3 threads*/
int numOfElements;/*Total # of elements from the user*/
struct Results{ /*Struct to hold the statistical results*/
int min;
int max;
int average;
}Results;
/*This function finds the minimum element of an array*/
void *findMin(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
Results.min = elements[0]; /*set minimum to first element */
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] < Results.min){ /*if the current element is less than the current min*/
Results.min = elements[i]; /*store the new min*/
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the maximum element of an array*/
void *findMax(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
if(elements[i] > Results.max){ /*store the new max*/
Results.max = elements[i];
}
}
running_threads -= 1; /*Decrement thread count*/
return NULL;
}
/*This function finds the average of an array*/
void *findAverage(void *array_ptr){
int i; /*counter*/
int *elements = (int*)array_ptr; /*re reference void array pointer*/
for(i = 0; i < numOfElements; i++){ /*iterate through array*/
Results.average += elements[i]; /*add element @ i to average*/
}
Results.average = Results.average/numOfElements; /*Divide the sum by the number of
elements*/
running_threads -= 1; /*Decrement running threads counter*/
return NULL;
}
/* This method accepts a int n(initial size of array) and
pointer to an array and returns # of elements in the array
*/
int getArrayInput(int n, int *array_ptr){
int input;/*Store user input */
int numberOfElements = 0;/*Number of Integers inputed*/
printf(\"Creating Dynamic Array...\ -\ \");
for(;;){ /*infinite loop*/
printf(\"Enter a positive value:\ Negative Number to Stop\ -\ \");
//Get Int from console, store at address of input
if (scanf(\"%d\",&input) != 1){
printf(\"\ Oops that wasn\'t an Integer\ lets try filling the array again\ Remember
INTEGERS only!\ \");
exit(EXIT_FAILURE);
}
if (input >= 0){
if (numberOfElements == n){
n += 1; //Make room for the current input
array_ptr = realloc(array_ptr, n * sizeof(int));//realloc array and set pointer
}
array_ptr[numberOfElements++] = input;//Store input at next empty element
} else {
printf(\"\ Number of Integers: %d\ \", numberOfElements);
break;
}
}
return numberOfElements;
}
/*This function joins our n number of threads */
void joinThreads(int numberOfThreads){
int i; /*count*/
int s; /*error #*/
while(numberOfThreads >= 0){ /*Join our threads*/
s = pthread_jo.
Note that there are two poles in the system. The common region of th.pdf
Note that there are two poles in the system. The common region of the entire system is inside the
one of the poles. Therefore, the region of convergence is less than the unit circle. Therefore, the
signal is left-sided.
If it is greater than unit circuit, it is a right-sided signal.
If it is between the poles, it is two-sided.
Solution
Note that there are two poles in the system. The common region of the entire system is inside the
one of the poles. Therefore, the region of convergence is less than the unit circle. Therefore, the
signal is left-sided.
If it is greater than unit circuit, it is a right-sided signal.
If it is between the poles, it is two-sided..
Obvious first to third shell. E is proportional .pdf
Obvious first to third shell. E is proportional to [1/n1^2 -1/n2^2] [1/n1^2 -1/n2^2]
is more when electron moves from the first to the third shel
Solution
Obvious first to third shell. E is proportional to [1/n1^2 -1/n2^2] [1/n1^2 -1/n2^2]
is more when electron moves from the first to the third shel.
a) YES. Both are hydrocarbons and nonpolar. Like .pdf
a) YES. Both are hydrocarbons and nonpolar. Like dissolves like. b) NO. CBr4 is
nonpolar and H2O is polar. Both do not mix. c) YES. LiNO3 is ionic and H2O is a polar solvent.
Polar solvents dissolve polar and ionic solutes. d) No. CH3OH is polar due to the O-H bond and
the other molecule is a hydrocarbon and nonpolar. They do not mix this would be helpful
Solution
a) YES. Both are hydrocarbons and nonpolar. Like dissolves like. b) NO. CBr4 is
nonpolar and H2O is polar. Both do not mix. c) YES. LiNO3 is ionic and H2O is a polar solvent.
Polar solvents dissolve polar and ionic solutes. d) No. CH3OH is polar due to the O-H bond and
the other molecule is a hydrocarbon and nonpolar. They do not mix this would be helpful.
Basic assumptions 1 Pairs of electrons in the va.pdf
Basic assumptions 1 Pairs of electrons in the valence shell of a central atom repel
each other. 2 These pairs of electrons tend to occupy positions in space that minimize repulsions
and maximize the distance of separation between them. 3 The valence shell is taken as a sphere
with electron pairs localizing on the spherical surface at maximum distance from one another. 4
A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a
multiple bond are treated as a single super pair. 5 Where two or more resonance structures can
depict a molecule the VSEPR model is applicable to any such structure.
Solution
Basic assumptions 1 Pairs of electrons in the valence shell of a central atom repel
each other. 2 These pairs of electrons tend to occupy positions in space that minimize repulsions
and maximize the distance of separation between them. 3 The valence shell is taken as a sphere
with electron pairs localizing on the spherical surface at maximum distance from one another. 4
A multiple bond is treated as if it is a single electron pair and the two or three electron pairs of a
multiple bond are treated as a single super pair. 5 Where two or more resonance structures can
depict a molecule the VSEPR model is applicable to any such structure..
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- 1. Option C Explanation: The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720. Solution Option C Explanation:
- 2. The word 'LEADING' has 7 different letters. When the vowels EAI are always together, they can be supposed to form one letter. Then, we have to arrange the letters LNDG (EAI). Now, 5 (4 + 1 = 5) letters can be arranged in 5! = 120 ways. The vowels (EAI) can be arranged among themselves in 3! = 6 ways. Required number of ways = (120 x 6) = 720.