Carbocations are stabilized by hyperconjugation b.pdf
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Carbocations are stabilized by hyperconjugation but only high order. Primary ones are NOT. They are highly unstable Canbanions, on the other hand are stabilized by conjugation. Resonance can stabilize these ions, esp if aromatically Solution Carbocations are stabilized by hyperconjugation but only high order. Primary ones are NOT. They are highly unstable Canbanions, on the other hand are stabilized by conjugation. Resonance can stabilize these ions, esp if aromatically.
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A is correct. Public and private community strings are used within S.pdf
A is correct. Public and private community strings are used within SNMP to read and write,
respectively.
B, C, and D are incorrect. B is incorrect because the choices are backwards. Private allows for
writing configuration data, while public allows for reading MIB information. C and D are
incorrect because they do not exist as community strings.
Solution
A is correct. Public and private community strings are used within SNMP to read and write,
respectively.
B, C, and D are incorrect. B is incorrect because the choices are backwards. Private allows for
writing configuration data, while public allows for reading MIB information. C and D are
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1. The smart phone, computer, navigation system and digital scanner .pdf
1. The smart phone, computer, navigation system and digital scanner have Microchip technology
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2.
3.
Solution
1. The smart phone, computer, navigation system and digital scanner have Microchip technology
in common.
2.
3..
1-They have terrestrial adaptation and dominant sporophytic phase..pdf
1-They have terrestrial adaptation and dominant sporophytic phase.
2-ovule is produced, but but it is not enclosed within ovary wall.
3- all are heterosporous(producing mega and microspores).microspore termed pollen grains
make direct contact with the ovules . microgametophytes develops within the pollenwall)i.e
endosporic) and without male sex organ, antheridium. megaspors bears megagametophyte which
contain multicellular female sex organ (archegonium).
4-they dont require water for sperm to swim in to reach the egg, all gymnosperm have non
flagellated sperms.
5-no double fertilization to produce endosperm . the endosperm nutritive tissue is haploid.
1-Dominant sporophytic phase.
2-ovule develops within ovary of a carpel, and enclosed by ovary wall.
3-All are heterosporous(producing mega and microspores).micro or male gametophytes develops
within microspore (pollengrain) and without antheridium.megaspore bear mega or female
gametophyes and without archegonium. the gametophytes of angiosperms are much reduced in
size and cell number. the mature microgametophytes within the pollen grain consist of only three
cells, and the mature megagametophytes within the ovule , in most angiosperm. consists of only
seven to eight nuclei.
4-pollination is indirect in which pollen falls on the stigma of the carpel rather than directly on
the ovules .
5-double fertilization occurs , to produce embryo as well as endosperm . the endosperm, nutritive
tissue for the future embryo , is not haploid but triploid and forms after
fertilization.gymnospermsAngiosperm
1-They have terrestrial adaptation and dominant sporophytic phase.
2-ovule is produced, but but it is not enclosed within ovary wall.
3- all are heterosporous(producing mega and microspores).microspore termed pollen grains
make direct contact with the ovules . microgametophytes develops within the pollenwall)i.e
endosporic) and without male sex organ, antheridium. megaspors bears megagametophyte
which contain multicellular female sex organ (archegonium).
4-they dont require water for sperm to swim in to reach the egg, all gymnosperm have non
flagellated sperms.
5-no double fertilization to produce endosperm . the endosperm nutritive tissue is haploid.
1-Dominant sporophytic phase.
2-ovule develops within ovary of a carpel, and enclosed by ovary wall.
3-All are heterosporous(producing mega and microspores).micro or male gametophytes
develops within microspore (pollengrain) and without antheridium.megaspore bear mega or
female gametophyes and without archegonium. the gametophytes of angiosperms are much
reduced in size and cell number. the mature microgametophytes within the pollen grain consist
of only three cells, and the mature megagametophytes within the ovule , in most angiosperm.
consists of only seven to eight nuclei.
4-pollination is indirect in which pollen falls on the stigma of the carpel rather than directly on
the ovules .
5-double fertilization occurs , to produce embryo as well as endosperm . th.
(1)Solution(1).pdf
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What does absense of MHC Class I indicateAlmost all nucleated cel.pdf
What does absense of MHC Class I indicate?
Almost all nucleated cells will have MHC Class I molecules on their surface and their main role
is to present foreign protein to cytotoxic T cells (Tc Cells) so that the cells which got infected
can be cleared by these Tc cells. But in some cases, it has been observed that some tumor cells
will escape killing or elimination by Tc cells by becoming MHC Class I negative, here they will
not express MHC Class I molecules on their surface and thus there will not be any expression of
tumor antigens to Tc cells and these will survive and will go on producing new cells and form
the tumor. Like this MHC class I negative cells present within the primary tumor will develop
into unnoticed micrometastases and will have an immunological equilibrium and succeeds in
surviving in a dormant stage for long periods of time
What does the presentation of tyrosinase indicate?
The tyrosinase will cleave the tumor proteins and will express the melanoma antigens on the cell
surface as peptides and overexpression of these antigens in tumor cells will result in an
abnormally high density of specific peptide:MHC complexes on the cell surface making them
immunogenic and provides them tolerance and is seen as tumor rejection in most of the cases.
What dos decreased CTL infiltratioin indicate?
It is the CD8+ cytotoxic T lymphocytes (CTLs) which will reject the tumor by infiltrating the
solid tumors by recognizing the tumor antigens, and will kill those tumor cells. And to do this
CTLs will try to infiltrate tumors in depth and it is done only when tumor cells express the
cognate CTL antigen. If the tumor will not express the cognate antigen, CTL infiltration will not
occur and the tumor cells cannot be killed and so the expression of tumor antigen on the cells
will attract CTL infiltration, at the same time when the tumor antigens are not presented on these
cells, it will decrease the CTL infiltration.
Solution
What does absense of MHC Class I indicate?
Almost all nucleated cells will have MHC Class I molecules on their surface and their main role
is to present foreign protein to cytotoxic T cells (Tc Cells) so that the cells which got infected
can be cleared by these Tc cells. But in some cases, it has been observed that some tumor cells
will escape killing or elimination by Tc cells by becoming MHC Class I negative, here they will
not express MHC Class I molecules on their surface and thus there will not be any expression of
tumor antigens to Tc cells and these will survive and will go on producing new cells and form
the tumor. Like this MHC class I negative cells present within the primary tumor will develop
into unnoticed micrometastases and will have an immunological equilibrium and succeeds in
surviving in a dormant stage for long periods of time
What does the presentation of tyrosinase indicate?
The tyrosinase will cleave the tumor proteins and will express the melanoma antigens on the cell
surface as peptides and overexpres.
Recommended
A is correct. Public and private community strings are used within S.pdf
A is correct. Public and private community strings are used within SNMP to read and write,
respectively.
B, C, and D are incorrect. B is incorrect because the choices are backwards. Private allows for
writing configuration data, while public allows for reading MIB information. C and D are
incorrect because they do not exist as community strings.
Solution
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respectively.
B, C, and D are incorrect. B is incorrect because the choices are backwards. Private allows for
writing configuration data, while public allows for reading MIB information. C and D are
incorrect because they do not exist as community strings..
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1. The smart phone, computer, navigation system and digital scanner have Microchip technology
in common.
2.
3.
Solution
1. The smart phone, computer, navigation system and digital scanner have Microchip technology
in common.
2.
3..
1-They have terrestrial adaptation and dominant sporophytic phase..pdf
1-They have terrestrial adaptation and dominant sporophytic phase.
2-ovule is produced, but but it is not enclosed within ovary wall.
3- all are heterosporous(producing mega and microspores).microspore termed pollen grains
make direct contact with the ovules . microgametophytes develops within the pollenwall)i.e
endosporic) and without male sex organ, antheridium. megaspors bears megagametophyte which
contain multicellular female sex organ (archegonium).
4-they dont require water for sperm to swim in to reach the egg, all gymnosperm have non
flagellated sperms.
5-no double fertilization to produce endosperm . the endosperm nutritive tissue is haploid.
1-Dominant sporophytic phase.
2-ovule develops within ovary of a carpel, and enclosed by ovary wall.
3-All are heterosporous(producing mega and microspores).micro or male gametophytes develops
within microspore (pollengrain) and without antheridium.megaspore bear mega or female
gametophyes and without archegonium. the gametophytes of angiosperms are much reduced in
size and cell number. the mature microgametophytes within the pollen grain consist of only three
cells, and the mature megagametophytes within the ovule , in most angiosperm. consists of only
seven to eight nuclei.
4-pollination is indirect in which pollen falls on the stigma of the carpel rather than directly on
the ovules .
5-double fertilization occurs , to produce embryo as well as endosperm . the endosperm, nutritive
tissue for the future embryo , is not haploid but triploid and forms after
fertilization.gymnospermsAngiosperm
1-They have terrestrial adaptation and dominant sporophytic phase.
2-ovule is produced, but but it is not enclosed within ovary wall.
3- all are heterosporous(producing mega and microspores).microspore termed pollen grains
make direct contact with the ovules . microgametophytes develops within the pollenwall)i.e
endosporic) and without male sex organ, antheridium. megaspors bears megagametophyte
which contain multicellular female sex organ (archegonium).
4-they dont require water for sperm to swim in to reach the egg, all gymnosperm have non
flagellated sperms.
5-no double fertilization to produce endosperm . the endosperm nutritive tissue is haploid.
1-Dominant sporophytic phase.
2-ovule develops within ovary of a carpel, and enclosed by ovary wall.
3-All are heterosporous(producing mega and microspores).micro or male gametophytes
develops within microspore (pollengrain) and without antheridium.megaspore bear mega or
female gametophyes and without archegonium. the gametophytes of angiosperms are much
reduced in size and cell number. the mature microgametophytes within the pollen grain consist
of only three cells, and the mature megagametophytes within the ovule , in most angiosperm.
consists of only seven to eight nuclei.
4-pollination is indirect in which pollen falls on the stigma of the carpel rather than directly on
the ovules .
5-double fertilization occurs , to produce embryo as well as endosperm . th.
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What does absense of MHC Class I indicateAlmost all nucleated cel.pdf
What does absense of MHC Class I indicate?
Almost all nucleated cells will have MHC Class I molecules on their surface and their main role
is to present foreign protein to cytotoxic T cells (Tc Cells) so that the cells which got infected
can be cleared by these Tc cells. But in some cases, it has been observed that some tumor cells
will escape killing or elimination by Tc cells by becoming MHC Class I negative, here they will
not express MHC Class I molecules on their surface and thus there will not be any expression of
tumor antigens to Tc cells and these will survive and will go on producing new cells and form
the tumor. Like this MHC class I negative cells present within the primary tumor will develop
into unnoticed micrometastases and will have an immunological equilibrium and succeeds in
surviving in a dormant stage for long periods of time
What does the presentation of tyrosinase indicate?
The tyrosinase will cleave the tumor proteins and will express the melanoma antigens on the cell
surface as peptides and overexpression of these antigens in tumor cells will result in an
abnormally high density of specific peptide:MHC complexes on the cell surface making them
immunogenic and provides them tolerance and is seen as tumor rejection in most of the cases.
What dos decreased CTL infiltratioin indicate?
It is the CD8+ cytotoxic T lymphocytes (CTLs) which will reject the tumor by infiltrating the
solid tumors by recognizing the tumor antigens, and will kill those tumor cells. And to do this
CTLs will try to infiltrate tumors in depth and it is done only when tumor cells express the
cognate CTL antigen. If the tumor will not express the cognate antigen, CTL infiltration will not
occur and the tumor cells cannot be killed and so the expression of tumor antigen on the cells
will attract CTL infiltration, at the same time when the tumor antigens are not presented on these
cells, it will decrease the CTL infiltration.
Solution
What does absense of MHC Class I indicate?
Almost all nucleated cells will have MHC Class I molecules on their surface and their main role
is to present foreign protein to cytotoxic T cells (Tc Cells) so that the cells which got infected
can be cleared by these Tc cells. But in some cases, it has been observed that some tumor cells
will escape killing or elimination by Tc cells by becoming MHC Class I negative, here they will
not express MHC Class I molecules on their surface and thus there will not be any expression of
tumor antigens to Tc cells and these will survive and will go on producing new cells and form
the tumor. Like this MHC class I negative cells present within the primary tumor will develop
into unnoticed micrometastases and will have an immunological equilibrium and succeeds in
surviving in a dormant stage for long periods of time
What does the presentation of tyrosinase indicate?
The tyrosinase will cleave the tumor proteins and will express the melanoma antigens on the cell
surface as peptides and overexpres.
The probabilities must add up to one. Add the ones you have and su.pdf
The probabilities must add up to one.
Add the ones you have and subtract their sum from one
to get the missing probability.
P(4) = 1- .74 = 0.26
Solution
The probabilities must add up to one.
Add the ones you have and subtract their sum from one
to get the missing probability.
P(4) = 1- .74 = 0.26.
The most likely F1 will be the one that is made from 4 and 5. This i.pdf
The document discusses two genetic maps between genes c and d. Map 1 shows a distance of 15 between the genes, while Map 2 shows a distance of 20. Since a greater distance between genes means more recombinants, the F1 offspring from parents 4 and 5 have a greater chance of losing homozygous recessive alleles due to less linkage between genes c and d in Map 2. Therefore, the F1 from parents 4 and 5 is most likely.
Subculturing or passage is the method of preserving or continuing th.pdf
Subculturing or passage is the method of preserving or continuing the culture for longer periods
Subculturing can be done in three procedures based on the cells we need to passage or
subculture.
1. Bacterial subculture.
2.Monolayer sub culture.
3. Suspension sub culture.
Requirements :-
For any of the sub culture it needs the following criteria
The adequate conditions like subculture should be done periodically, need of adding nutrients or
shifting of the innoculum both are considered as subculture.
1. MONOLAYER CULTURES when the continous cell lines or other adhered cells which are
grown on the roux flask or other surface which are cultured by removing by trypsinization
procedures and grown in the fresh medium eg DMEM medium Dulbeco modified Eagles
medium. So that the cells will grow vigorously and differentiate.
2. Suspension cultures when cells of the well grown medium is transferred to the fresh medium
components so that cells grow in high nutrient concentration,
3. Bacterial sub culturing : Transfer of innoculum(colony) from solid agar surface to broth
medium, solid surface to solid surface, liquid broth to liquid broth, liquid broth to solid agar
medium surface, sub culturing can be done by streaking, pouring, spreading onto fresh medium
components.
Solution
Subculturing or passage is the method of preserving or continuing the culture for longer periods
Subculturing can be done in three procedures based on the cells we need to passage or
subculture.
1. Bacterial subculture.
2.Monolayer sub culture.
3. Suspension sub culture.
Requirements :-
For any of the sub culture it needs the following criteria
The adequate conditions like subculture should be done periodically, need of adding nutrients or
shifting of the innoculum both are considered as subculture.
1. MONOLAYER CULTURES when the continous cell lines or other adhered cells which are
grown on the roux flask or other surface which are cultured by removing by trypsinization
procedures and grown in the fresh medium eg DMEM medium Dulbeco modified Eagles
medium. So that the cells will grow vigorously and differentiate.
2. Suspension cultures when cells of the well grown medium is transferred to the fresh medium
components so that cells grow in high nutrient concentration,
3. Bacterial sub culturing : Transfer of innoculum(colony) from solid agar surface to broth
medium, solid surface to solid surface, liquid broth to liquid broth, liquid broth to solid agar
medium surface, sub culturing can be done by streaking, pouring, spreading onto fresh medium
components..
public class Interest{double interest(double rate, double amount){.pdf
public class Interest{
double interest(double rate, double amount){
return amount+(amount*rate);
}
double totalSaving(int months,double monthly,double rate){
double yearly=rate/12;
for(int i=1;i<=months;i++){
System.out.print(\"After Month \"+i+\" Amount \");
monthly=interest(yearly,monthly);
System.out.println(monthly);
}
}
public static void main(String [] args){
Interest i=new Interest();
i.totalSaving(12,100,0.5)
}
}
Solution
public class Interest{
double interest(double rate, double amount){
return amount+(amount*rate);
}
double totalSaving(int months,double monthly,double rate){
double yearly=rate/12;
for(int i=1;i<=months;i++){
System.out.print(\"After Month \"+i+\" Amount \");
monthly=interest(yearly,monthly);
System.out.println(monthly);
}
}
public static void main(String [] args){
Interest i=new Interest();
i.totalSaving(12,100,0.5)
}
}.
Phototaxisthe Phenomenon of movement of motile unicellular organis.pdf
Phototaxis
the Phenomenon of movement of motile unicellular organisms such as algae and bacteria toward
or away from light is termed as phototaxis movement.
This is advantageous for phototrophic organisms as they can orient themselves most efficiently
to receive light for photosynthesis.
Phototaxis is called positive if the movement is in the direction of light and negative if the
direction is opposite.
Phototropism,
Growth with respect to light, expressed in all shoots and some roots.
Phototropism is one of the many plant tropisms or movements which respond to external stimuli
Charles Darwin studied plant growth phenomena involving tropisms.
He found the bending of plants toward light. This phenomenon, which is caused by differential
growth, is called phototropism.
In experiments he used seedlings of canary grass (Phalaris canariensis), the youngest leaves are
sheathed in a protective organ called the coleoptile.
Coleoptiles are very sensitive to light, especially to blue light . If illuminated on one side with a
short pulse of dim blue light, they will bend (grow) toward the source of the light pulse within an
hour.
Simillarity between Phototaxis and phototropism
Both are related with abiotic factor i.e light.
___________________. .
Solution
Phototaxis
the Phenomenon of movement of motile unicellular organisms such as algae and bacteria toward
or away from light is termed as phototaxis movement.
This is advantageous for phototrophic organisms as they can orient themselves most efficiently
to receive light for photosynthesis.
Phototaxis is called positive if the movement is in the direction of light and negative if the
direction is opposite.
Phototropism,
Growth with respect to light, expressed in all shoots and some roots.
Phototropism is one of the many plant tropisms or movements which respond to external stimuli
Charles Darwin studied plant growth phenomena involving tropisms.
He found the bending of plants toward light. This phenomenon, which is caused by differential
growth, is called phototropism.
In experiments he used seedlings of canary grass (Phalaris canariensis), the youngest leaves are
sheathed in a protective organ called the coleoptile.
Coleoptiles are very sensitive to light, especially to blue light . If illuminated on one side with a
short pulse of dim blue light, they will bend (grow) toward the source of the light pulse within an
hour.
Simillarity between Phototaxis and phototropism
Both are related with abiotic factor i.e light.
___________________. ..
Miller indices It is defined as by considerin.pdf
Miller indices:
It is defined as by considering the plane in which it intersects the main
crystallographic axes of the solid. A set of numbers which explains the intercepts and thus it may
be used to identify the plane or the surface.
Solution
Miller indices:
It is defined as by considering the plane in which it intersects the main
crystallographic axes of the solid. A set of numbers which explains the intercepts and thus it may
be used to identify the plane or the surface..
Marijuana is also called as cannabis and it is a preparation of cann.pdf
Marijuana is also called as cannabis and it is a preparation of cannabis plant and it is used as a
psychoactive drug or medicine.
Characteristics of Cannabis were studied to determine the suitable and efficient environmental
conditions for its indoor mass cultivation for pharmaceutical studies.
Effect of different photosynthetic photon flux densities,temperature and CO2 concentrations on
gas and water vapour exchange characteristics of the Cannabis.
The rate of photosynthesis and water use efficiency of Cannabis increases with photosynthetic
photon flux densities at lower temperatures.
Genrally the photosynthetic process will occur by taking the carbondioxide and water and in the
presence of sunlight and chlorophyll it forms the glucose and oxygen and from the glucose the
plant makes the cellulose which gives the fibrous structures to the plant.
The osmosis process helps in the absorbption of water and minerals from soil. The osmotic
pressure mainly helpful in intake and transport of minerals and the soil condition also plays an
important role.
Cannabis gives its potential for better survival,growth and productivity in drier and CO2 rich
environment.
Solution
Marijuana is also called as cannabis and it is a preparation of cannabis plant and it is used as a
psychoactive drug or medicine.
Characteristics of Cannabis were studied to determine the suitable and efficient environmental
conditions for its indoor mass cultivation for pharmaceutical studies.
Effect of different photosynthetic photon flux densities,temperature and CO2 concentrations on
gas and water vapour exchange characteristics of the Cannabis.
The rate of photosynthesis and water use efficiency of Cannabis increases with photosynthetic
photon flux densities at lower temperatures.
Genrally the photosynthetic process will occur by taking the carbondioxide and water and in the
presence of sunlight and chlorophyll it forms the glucose and oxygen and from the glucose the
plant makes the cellulose which gives the fibrous structures to the plant.
The osmosis process helps in the absorbption of water and minerals from soil. The osmotic
pressure mainly helpful in intake and transport of minerals and the soil condition also plays an
important role.
Cannabis gives its potential for better survival,growth and productivity in drier and CO2 rich
environment..
IHD = 1 Therefore there can be only 1 -bond or ring. The two signals.pdf
IHD = 1 Therefore there can be only 1 -bond or ring. The two signals composed of singlets
means all the hydrogens are isolated from one another, but there are two types in the molecule
giving differing chemical shifts. The structure must be symmetrical. Likely compound is 2,2-
dimethylpropanal (three methyl groups contributing one signal, and one aldehyde hydrogen
contributing another signal. All non-equivalent hydrogens are too far apart to split the signal,
resulting in singlets)
Solution
IHD = 1 Therefore there can be only 1 -bond or ring. The two signals composed of singlets
means all the hydrogens are isolated from one another, but there are two types in the molecule
giving differing chemical shifts. The structure must be symmetrical. Likely compound is 2,2-
dimethylpropanal (three methyl groups contributing one signal, and one aldehyde hydrogen
contributing another signal. All non-equivalent hydrogens are too far apart to split the signal,
resulting in singlets).
I have the answer ready in paint but i am not able to upload. 1) F.pdf
I have the answer ready in paint but i am not able to upload.
1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0)
Hence the and moves in a circle of radius 1 unit.
2) F\'(t)= (-2sin2t, 2cos2t)
F\'(pi/2) = (0, -2)
This represents the Speed at t= pi/2
3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1)
Solution
I have the answer ready in paint but i am not able to upload.
1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0)
Hence the and moves in a circle of radius 1 unit.
2) F\'(t)= (-2sin2t, 2cos2t)
F\'(pi/2) = (0, -2)
This represents the Speed at t= pi/2
3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1).
homedesigninsertanimationsdevelopmentSolutionhomedes.pdf
This document appears to be about home design and development. It mentions home design, insert animations, and development as a potential solution. The document repeats these terms but does not provide much additional context or details to understand its purpose or meaning.
C is correct. DHCP is a statefull method of configuring IPv6 address.pdf
C is correct. DHCP is a statefull method of configuring IPv6 addresses. It maintains the address
states of each device.
A, B, and D are incorrect. Each is a stateless method with no device maintaining the address
states.
Solution
C is correct. DHCP is a statefull method of configuring IPv6 addresses. It maintains the address
states of each device.
A, B, and D are incorrect. Each is a stateless method with no device maintaining the address
states..
B) The energy required to remove an electron from an atomSolutio.pdf
B) The energy required to remove an electron from an atom
Solution
B) The energy required to remove an electron from an atom.
AnswerThe following molecules would not require carrier transport.pdf
Answer:
The following molecules would not require carrier transporters:
a) carbon dioxide
c) progesterone
The following molecules would require carrier transporters:
b) fructose
d) arganine
e) adenosine diphosphate
Polar molecules and large ions dissolved in water cannot diffuse freely across the plasma
membrane due to the hydrophobic nature of the fatty acid tails of the phospholipids that make up
the lipid bilayer. Only small, non-polar molecules, such as oxygen and carbon dioxide, can
diffuse easily across the membrane. Hence, no nonpolar molecules are transported by proteins in
the form of transmembrane channels. These channels are gated, meaning that they open and
close, and thus deregulate the flow of ions or small polar molecules across membranes,
sometimes against the osmotic gradient. Larger molecules are transported by transmembrane
carrier proteins, such as permeases, that change their conformation as the molecules are carried
across (e.g. glucose or amino acids). Non-polar molecules, such as retinol or lipids, are poorly
soluble in water. They are transported through aqueous compartments of cells or through
extracellular space by water-soluble carriers (e.g. retinol binding protein).
ADP/ATP translocases, also known as adenine nucleotide translocases (ANT) and ADP/ATP
carrier proteins (AAC), are transporter proteins that enable the exchange of cytosolic adenosine
diphosphate(ADP) and mitochondrial adenosine triphosphate (ATP) across the inner
mitochondrial membrane.
Solution
Answer:
The following molecules would not require carrier transporters:
a) carbon dioxide
c) progesterone
The following molecules would require carrier transporters:
b) fructose
d) arganine
e) adenosine diphosphate
Polar molecules and large ions dissolved in water cannot diffuse freely across the plasma
membrane due to the hydrophobic nature of the fatty acid tails of the phospholipids that make up
the lipid bilayer. Only small, non-polar molecules, such as oxygen and carbon dioxide, can
diffuse easily across the membrane. Hence, no nonpolar molecules are transported by proteins in
the form of transmembrane channels. These channels are gated, meaning that they open and
close, and thus deregulate the flow of ions or small polar molecules across membranes,
sometimes against the osmotic gradient. Larger molecules are transported by transmembrane
carrier proteins, such as permeases, that change their conformation as the molecules are carried
across (e.g. glucose or amino acids). Non-polar molecules, such as retinol or lipids, are poorly
soluble in water. They are transported through aqueous compartments of cells or through
extracellular space by water-soluble carriers (e.g. retinol binding protein).
ADP/ATP translocases, also known as adenine nucleotide translocases (ANT) and ADP/ATP
carrier proteins (AAC), are transporter proteins that enable the exchange of cytosolic adenosine
diphosphate(ADP) and mitochondrial adenosine triphosphate (ATP) across the inner
mitoc.
AnswerCyanide is an irreversible enzyme inhibitor of Cytochrome C.pdf
Answer:
Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth
complex in the electron transport chain.
Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which
results in blockade of electrons transfer to the final electron acceptor oxygen.
This blockade causes build up of proton motive force in the intermembrane space as the protons
are not allowed to flow back into the matrix which subsequently halts ATP production.
Solution
Answer:
Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth
complex in the electron transport chain.
Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which
results in blockade of electrons transfer to the final electron acceptor oxygen.
This blockade causes build up of proton motive force in the intermembrane space as the protons
are not allowed to flow back into the matrix which subsequently halts ATP production..
a.There are two variables in labor market model. One is wage rate .pdf
a.
There are two variables in labor market model. One is wage rate and the other is labor quantity.
The ‘X’ axis represents labor quantity and the ‘Y’ axis represents wage rate.
Solution
a.
There are two variables in labor market model. One is wage rate and the other is labor quantity.
The ‘X’ axis represents labor quantity and the ‘Y’ axis represents wage rate..
A respiratory therapy professional diagnoses and treats those suffer.pdf
A respiratory therapy professional diagnoses and treats those suffering from chronic and other
respiratory problems. Some of their patients with respiratory problems would be:
• Premature infants
• Asthma patients
• Patients with lung infections
• Heart patients
A respiratory therapist (RT) has a variety of job responsibilities. These include talking to
patients, gathering information regarding their medical condition, diagnosing breathing
conditions and associated treatments, informing the family about the respiratory therapy
treatments, etc. Respiratory therapist must update their professional respiratory knowledge
periodically and need to be well versed with the latest available techniques, treatments and
diseases in respiratory therapy.
Care
Respiratory therapists often work in medical facilities such as hospitals and nursing homes. They
administer oxygen, perform CPR, use ventilators, and give medicines as needed. The ability to
provide this care and build relationships with patients is an advantage for people who want to
work in the medical field without becoming a medical doctor. Therapists who work in extended
care facilities like nursing homes have an even longer time to develop relationships with their
patients.
Misconception among nurses about a respiration therapist:
STRESS:
Nurses do think that, being a respiration therapist could be strenuous and stressful, as they spend
the majority of the day on their feet.
The care that respiratory therapists provide deals with one of the most basic and important life
activities -- breathing. Helping people through breathing struggles is stressful under normal
conditions. For those therapists who work in hospitals, the stress is more constant. Administering
CPR and assisting with treatment in emergency rooms is high pressure.
Another factor that can add to the stress is working with children who have breathing issues, or
who have been in a drowning accident or other type of accident that has compromised their
respiration.
WORK CONDITION and ENVIRONMENT:
Nurses do believe that, RTs may have to work on weekends, holidays, and various shifts and
may also have to work extended hours in case of emergency. A career in respiratory therapy also
means being exposed to infections and micro-organisms.
Another misconception is that, as being a registered nurse, they do have more opportunities of
different places to work like hospitals, clinics, doctor\'s offices, jail\'s, factories, home health, and
of course as a school nurse. whereas, as RTs, well not so many as they could work on just
hospitals and home health.
These are the major misconceptions among nurses about a respiratory therapist.
Solution
A respiratory therapy professional diagnoses and treats those suffering from chronic and other
respiratory problems. Some of their patients with respiratory problems would be:
• Premature infants
• Asthma patients
• Patients with lung infections
• Heart patients
A respiratory therapist (RT) has a variety of .
a) Angular gyrus is present in parietal cortex. It is concerned with.pdf
a) Angular gyrus is present in parietal cortex. It is concerned with storage of memories of the
visual symbols.
b) Broca\'s aphasia occurs due to damage of left inferior frontal region of cerebral cortex.
Broca’s area:
It is area 44. Area 44 is the motor speech area. It is also called speech center. It is situated in the
lower part of prefrontal cortex.
Broca’s aphasia:
c) Wernicke\'s area is situated in upper part of temporal lobe. It plays an important role in
speech. Lesions in this area results in Wernicke\'s aphasia.
d) Damage to left temporal lobe leads to Wernicke\'s aphasia.
Therefore, option b is correct.
Broca’s aphasia is usually associated with the lesions of left inferior frontal region of cerebral
cortex.
Solution
a) Angular gyrus is present in parietal cortex. It is concerned with storage of memories of the
visual symbols.
b) Broca\'s aphasia occurs due to damage of left inferior frontal region of cerebral cortex.
Broca’s area:
It is area 44. Area 44 is the motor speech area. It is also called speech center. It is situated in the
lower part of prefrontal cortex.
Broca’s aphasia:
c) Wernicke\'s area is situated in upper part of temporal lobe. It plays an important role in
speech. Lesions in this area results in Wernicke\'s aphasia.
d) Damage to left temporal lobe leads to Wernicke\'s aphasia.
Therefore, option b is correct.
Broca’s aphasia is usually associated with the lesions of left inferior frontal region of cerebral
cortex..
Since youre diluting the solution to 2.5 times .pdf
Since you\'re diluting the solution to 2.5 times its initial volume, since VM =
constant, the final concentration is the initial concentration over 2.5, or .0250/2.5 = .010 M.
Solution
Since you\'re diluting the solution to 2.5 times its initial volume, since VM =
constant, the final concentration is the initial concentration over 2.5, or .0250/2.5 = .010 M..
wo (-2) above the O3 means there are 2 extra elec.pdf
wo (-2) above the O3 means there are 2 extra electrons available for bonding.
Polyatomic ions are groups of atoms joined together by covalent bonds, like molecules, except
that they have a charge. The only difference between the process for drawing Lewis diagrams for
molecules and for ions is in determining the number of available electrons. Count the valence as
before then add an electron for each unit of negative charge or subtract one for each unit of
positive charge. For example, the sulfite ion has the formula SO32-. The number of valence
electrons is 6 + 3*6 + 2 = 26. The atom with the highest electronegativity in the sulfite ion is the
sulfur, so it belongs in the center of the diagram with oxygen atoms bonded to it. Fill in enough
bonds to hold the ion together then add non-bonding electrons until all atoms obey the octet rule
or you have used all that are available.
Solution
wo (-2) above the O3 means there are 2 extra electrons available for bonding.
Polyatomic ions are groups of atoms joined together by covalent bonds, like molecules, except
that they have a charge. The only difference between the process for drawing Lewis diagrams for
molecules and for ions is in determining the number of available electrons. Count the valence as
before then add an electron for each unit of negative charge or subtract one for each unit of
positive charge. For example, the sulfite ion has the formula SO32-. The number of valence
electrons is 6 + 3*6 + 2 = 26. The atom with the highest electronegativity in the sulfite ion is the
sulfur, so it belongs in the center of the diagram with oxygen atoms bonded to it. Fill in enough
bonds to hold the ion together then add non-bonding electrons until all atoms obey the octet rule
or you have used all that are available..
Gender and Mental Health - Counselling and Family Therapy Applications and In...
A proprietary approach developed by bringing together the best of learning theories from Psychology, design principles from the world of visualization, and pedagogical methods from over a decade of training experience, that enables you to: Learn better, faster!
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More from Ankitagarwaleleraipu
The probabilities must add up to one. Add the ones you have and su.pdf
The probabilities must add up to one.
Add the ones you have and subtract their sum from one
to get the missing probability.
P(4) = 1- .74 = 0.26
Solution
The probabilities must add up to one.
Add the ones you have and subtract their sum from one
to get the missing probability.
P(4) = 1- .74 = 0.26.
The most likely F1 will be the one that is made from 4 and 5. This i.pdf
The document discusses two genetic maps between genes c and d. Map 1 shows a distance of 15 between the genes, while Map 2 shows a distance of 20. Since a greater distance between genes means more recombinants, the F1 offspring from parents 4 and 5 have a greater chance of losing homozygous recessive alleles due to less linkage between genes c and d in Map 2. Therefore, the F1 from parents 4 and 5 is most likely.
Subculturing or passage is the method of preserving or continuing th.pdf
Subculturing or passage is the method of preserving or continuing the culture for longer periods
Subculturing can be done in three procedures based on the cells we need to passage or
subculture.
1. Bacterial subculture.
2.Monolayer sub culture.
3. Suspension sub culture.
Requirements :-
For any of the sub culture it needs the following criteria
The adequate conditions like subculture should be done periodically, need of adding nutrients or
shifting of the innoculum both are considered as subculture.
1. MONOLAYER CULTURES when the continous cell lines or other adhered cells which are
grown on the roux flask or other surface which are cultured by removing by trypsinization
procedures and grown in the fresh medium eg DMEM medium Dulbeco modified Eagles
medium. So that the cells will grow vigorously and differentiate.
2. Suspension cultures when cells of the well grown medium is transferred to the fresh medium
components so that cells grow in high nutrient concentration,
3. Bacterial sub culturing : Transfer of innoculum(colony) from solid agar surface to broth
medium, solid surface to solid surface, liquid broth to liquid broth, liquid broth to solid agar
medium surface, sub culturing can be done by streaking, pouring, spreading onto fresh medium
components.
Solution
Subculturing or passage is the method of preserving or continuing the culture for longer periods
Subculturing can be done in three procedures based on the cells we need to passage or
subculture.
1. Bacterial subculture.
2.Monolayer sub culture.
3. Suspension sub culture.
Requirements :-
For any of the sub culture it needs the following criteria
The adequate conditions like subculture should be done periodically, need of adding nutrients or
shifting of the innoculum both are considered as subculture.
1. MONOLAYER CULTURES when the continous cell lines or other adhered cells which are
grown on the roux flask or other surface which are cultured by removing by trypsinization
procedures and grown in the fresh medium eg DMEM medium Dulbeco modified Eagles
medium. So that the cells will grow vigorously and differentiate.
2. Suspension cultures when cells of the well grown medium is transferred to the fresh medium
components so that cells grow in high nutrient concentration,
3. Bacterial sub culturing : Transfer of innoculum(colony) from solid agar surface to broth
medium, solid surface to solid surface, liquid broth to liquid broth, liquid broth to solid agar
medium surface, sub culturing can be done by streaking, pouring, spreading onto fresh medium
components..
public class Interest{double interest(double rate, double amount){.pdf
public class Interest{
double interest(double rate, double amount){
return amount+(amount*rate);
}
double totalSaving(int months,double monthly,double rate){
double yearly=rate/12;
for(int i=1;i<=months;i++){
System.out.print(\"After Month \"+i+\" Amount \");
monthly=interest(yearly,monthly);
System.out.println(monthly);
}
}
public static void main(String [] args){
Interest i=new Interest();
i.totalSaving(12,100,0.5)
}
}
Solution
public class Interest{
double interest(double rate, double amount){
return amount+(amount*rate);
}
double totalSaving(int months,double monthly,double rate){
double yearly=rate/12;
for(int i=1;i<=months;i++){
System.out.print(\"After Month \"+i+\" Amount \");
monthly=interest(yearly,monthly);
System.out.println(monthly);
}
}
public static void main(String [] args){
Interest i=new Interest();
i.totalSaving(12,100,0.5)
}
}.
Phototaxisthe Phenomenon of movement of motile unicellular organis.pdf
Phototaxis
the Phenomenon of movement of motile unicellular organisms such as algae and bacteria toward
or away from light is termed as phototaxis movement.
This is advantageous for phototrophic organisms as they can orient themselves most efficiently
to receive light for photosynthesis.
Phototaxis is called positive if the movement is in the direction of light and negative if the
direction is opposite.
Phototropism,
Growth with respect to light, expressed in all shoots and some roots.
Phototropism is one of the many plant tropisms or movements which respond to external stimuli
Charles Darwin studied plant growth phenomena involving tropisms.
He found the bending of plants toward light. This phenomenon, which is caused by differential
growth, is called phototropism.
In experiments he used seedlings of canary grass (Phalaris canariensis), the youngest leaves are
sheathed in a protective organ called the coleoptile.
Coleoptiles are very sensitive to light, especially to blue light . If illuminated on one side with a
short pulse of dim blue light, they will bend (grow) toward the source of the light pulse within an
hour.
Simillarity between Phototaxis and phototropism
Both are related with abiotic factor i.e light.
___________________. .
Solution
Phototaxis
the Phenomenon of movement of motile unicellular organisms such as algae and bacteria toward
or away from light is termed as phototaxis movement.
This is advantageous for phototrophic organisms as they can orient themselves most efficiently
to receive light for photosynthesis.
Phototaxis is called positive if the movement is in the direction of light and negative if the
direction is opposite.
Phototropism,
Growth with respect to light, expressed in all shoots and some roots.
Phototropism is one of the many plant tropisms or movements which respond to external stimuli
Charles Darwin studied plant growth phenomena involving tropisms.
He found the bending of plants toward light. This phenomenon, which is caused by differential
growth, is called phototropism.
In experiments he used seedlings of canary grass (Phalaris canariensis), the youngest leaves are
sheathed in a protective organ called the coleoptile.
Coleoptiles are very sensitive to light, especially to blue light . If illuminated on one side with a
short pulse of dim blue light, they will bend (grow) toward the source of the light pulse within an
hour.
Simillarity between Phototaxis and phototropism
Both are related with abiotic factor i.e light.
___________________. ..
Miller indices It is defined as by considerin.pdf
Miller indices:
It is defined as by considering the plane in which it intersects the main
crystallographic axes of the solid. A set of numbers which explains the intercepts and thus it may
be used to identify the plane or the surface.
Solution
Miller indices:
It is defined as by considering the plane in which it intersects the main
crystallographic axes of the solid. A set of numbers which explains the intercepts and thus it may
be used to identify the plane or the surface..
Marijuana is also called as cannabis and it is a preparation of cann.pdf
Marijuana is also called as cannabis and it is a preparation of cannabis plant and it is used as a
psychoactive drug or medicine.
Characteristics of Cannabis were studied to determine the suitable and efficient environmental
conditions for its indoor mass cultivation for pharmaceutical studies.
Effect of different photosynthetic photon flux densities,temperature and CO2 concentrations on
gas and water vapour exchange characteristics of the Cannabis.
The rate of photosynthesis and water use efficiency of Cannabis increases with photosynthetic
photon flux densities at lower temperatures.
Genrally the photosynthetic process will occur by taking the carbondioxide and water and in the
presence of sunlight and chlorophyll it forms the glucose and oxygen and from the glucose the
plant makes the cellulose which gives the fibrous structures to the plant.
The osmosis process helps in the absorbption of water and minerals from soil. The osmotic
pressure mainly helpful in intake and transport of minerals and the soil condition also plays an
important role.
Cannabis gives its potential for better survival,growth and productivity in drier and CO2 rich
environment.
Solution
Marijuana is also called as cannabis and it is a preparation of cannabis plant and it is used as a
psychoactive drug or medicine.
Characteristics of Cannabis were studied to determine the suitable and efficient environmental
conditions for its indoor mass cultivation for pharmaceutical studies.
Effect of different photosynthetic photon flux densities,temperature and CO2 concentrations on
gas and water vapour exchange characteristics of the Cannabis.
The rate of photosynthesis and water use efficiency of Cannabis increases with photosynthetic
photon flux densities at lower temperatures.
Genrally the photosynthetic process will occur by taking the carbondioxide and water and in the
presence of sunlight and chlorophyll it forms the glucose and oxygen and from the glucose the
plant makes the cellulose which gives the fibrous structures to the plant.
The osmosis process helps in the absorbption of water and minerals from soil. The osmotic
pressure mainly helpful in intake and transport of minerals and the soil condition also plays an
important role.
Cannabis gives its potential for better survival,growth and productivity in drier and CO2 rich
environment..
IHD = 1 Therefore there can be only 1 -bond or ring. The two signals.pdf
IHD = 1 Therefore there can be only 1 -bond or ring. The two signals composed of singlets
means all the hydrogens are isolated from one another, but there are two types in the molecule
giving differing chemical shifts. The structure must be symmetrical. Likely compound is 2,2-
dimethylpropanal (three methyl groups contributing one signal, and one aldehyde hydrogen
contributing another signal. All non-equivalent hydrogens are too far apart to split the signal,
resulting in singlets)
Solution
IHD = 1 Therefore there can be only 1 -bond or ring. The two signals composed of singlets
means all the hydrogens are isolated from one another, but there are two types in the molecule
giving differing chemical shifts. The structure must be symmetrical. Likely compound is 2,2-
dimethylpropanal (three methyl groups contributing one signal, and one aldehyde hydrogen
contributing another signal. All non-equivalent hydrogens are too far apart to split the signal,
resulting in singlets).
I have the answer ready in paint but i am not able to upload. 1) F.pdf
I have the answer ready in paint but i am not able to upload.
1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0)
Hence the and moves in a circle of radius 1 unit.
2) F\'(t)= (-2sin2t, 2cos2t)
F\'(pi/2) = (0, -2)
This represents the Speed at t= pi/2
3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1)
Solution
I have the answer ready in paint but i am not able to upload.
1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0)
Hence the and moves in a circle of radius 1 unit.
2) F\'(t)= (-2sin2t, 2cos2t)
F\'(pi/2) = (0, -2)
This represents the Speed at t= pi/2
3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1).
homedesigninsertanimationsdevelopmentSolutionhomedes.pdf
This document appears to be about home design and development. It mentions home design, insert animations, and development as a potential solution. The document repeats these terms but does not provide much additional context or details to understand its purpose or meaning.
C is correct. DHCP is a statefull method of configuring IPv6 address.pdf
C is correct. DHCP is a statefull method of configuring IPv6 addresses. It maintains the address
states of each device.
A, B, and D are incorrect. Each is a stateless method with no device maintaining the address
states.
Solution
C is correct. DHCP is a statefull method of configuring IPv6 addresses. It maintains the address
states of each device.
A, B, and D are incorrect. Each is a stateless method with no device maintaining the address
states..
B) The energy required to remove an electron from an atomSolutio.pdf
B) The energy required to remove an electron from an atom
Solution
B) The energy required to remove an electron from an atom.
AnswerThe following molecules would not require carrier transport.pdf
Answer:
The following molecules would not require carrier transporters:
a) carbon dioxide
c) progesterone
The following molecules would require carrier transporters:
b) fructose
d) arganine
e) adenosine diphosphate
Polar molecules and large ions dissolved in water cannot diffuse freely across the plasma
membrane due to the hydrophobic nature of the fatty acid tails of the phospholipids that make up
the lipid bilayer. Only small, non-polar molecules, such as oxygen and carbon dioxide, can
diffuse easily across the membrane. Hence, no nonpolar molecules are transported by proteins in
the form of transmembrane channels. These channels are gated, meaning that they open and
close, and thus deregulate the flow of ions or small polar molecules across membranes,
sometimes against the osmotic gradient. Larger molecules are transported by transmembrane
carrier proteins, such as permeases, that change their conformation as the molecules are carried
across (e.g. glucose or amino acids). Non-polar molecules, such as retinol or lipids, are poorly
soluble in water. They are transported through aqueous compartments of cells or through
extracellular space by water-soluble carriers (e.g. retinol binding protein).
ADP/ATP translocases, also known as adenine nucleotide translocases (ANT) and ADP/ATP
carrier proteins (AAC), are transporter proteins that enable the exchange of cytosolic adenosine
diphosphate(ADP) and mitochondrial adenosine triphosphate (ATP) across the inner
mitochondrial membrane.
Solution
Answer:
The following molecules would not require carrier transporters:
a) carbon dioxide
c) progesterone
The following molecules would require carrier transporters:
b) fructose
d) arganine
e) adenosine diphosphate
Polar molecules and large ions dissolved in water cannot diffuse freely across the plasma
membrane due to the hydrophobic nature of the fatty acid tails of the phospholipids that make up
the lipid bilayer. Only small, non-polar molecules, such as oxygen and carbon dioxide, can
diffuse easily across the membrane. Hence, no nonpolar molecules are transported by proteins in
the form of transmembrane channels. These channels are gated, meaning that they open and
close, and thus deregulate the flow of ions or small polar molecules across membranes,
sometimes against the osmotic gradient. Larger molecules are transported by transmembrane
carrier proteins, such as permeases, that change their conformation as the molecules are carried
across (e.g. glucose or amino acids). Non-polar molecules, such as retinol or lipids, are poorly
soluble in water. They are transported through aqueous compartments of cells or through
extracellular space by water-soluble carriers (e.g. retinol binding protein).
ADP/ATP translocases, also known as adenine nucleotide translocases (ANT) and ADP/ATP
carrier proteins (AAC), are transporter proteins that enable the exchange of cytosolic adenosine
diphosphate(ADP) and mitochondrial adenosine triphosphate (ATP) across the inner
mitoc.
AnswerCyanide is an irreversible enzyme inhibitor of Cytochrome C.pdf
Answer:
Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth
complex in the electron transport chain.
Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which
results in blockade of electrons transfer to the final electron acceptor oxygen.
This blockade causes build up of proton motive force in the intermembrane space as the protons
are not allowed to flow back into the matrix which subsequently halts ATP production.
Solution
Answer:
Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth
complex in the electron transport chain.
Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which
results in blockade of electrons transfer to the final electron acceptor oxygen.
This blockade causes build up of proton motive force in the intermembrane space as the protons
are not allowed to flow back into the matrix which subsequently halts ATP production..
a.There are two variables in labor market model. One is wage rate .pdf
a.
There are two variables in labor market model. One is wage rate and the other is labor quantity.
The ‘X’ axis represents labor quantity and the ‘Y’ axis represents wage rate.
Solution
a.
There are two variables in labor market model. One is wage rate and the other is labor quantity.
The ‘X’ axis represents labor quantity and the ‘Y’ axis represents wage rate..
A respiratory therapy professional diagnoses and treats those suffer.pdf
A respiratory therapy professional diagnoses and treats those suffering from chronic and other
respiratory problems. Some of their patients with respiratory problems would be:
• Premature infants
• Asthma patients
• Patients with lung infections
• Heart patients
A respiratory therapist (RT) has a variety of job responsibilities. These include talking to
patients, gathering information regarding their medical condition, diagnosing breathing
conditions and associated treatments, informing the family about the respiratory therapy
treatments, etc. Respiratory therapist must update their professional respiratory knowledge
periodically and need to be well versed with the latest available techniques, treatments and
diseases in respiratory therapy.
Care
Respiratory therapists often work in medical facilities such as hospitals and nursing homes. They
administer oxygen, perform CPR, use ventilators, and give medicines as needed. The ability to
provide this care and build relationships with patients is an advantage for people who want to
work in the medical field without becoming a medical doctor. Therapists who work in extended
care facilities like nursing homes have an even longer time to develop relationships with their
patients.
Misconception among nurses about a respiration therapist:
STRESS:
Nurses do think that, being a respiration therapist could be strenuous and stressful, as they spend
the majority of the day on their feet.
The care that respiratory therapists provide deals with one of the most basic and important life
activities -- breathing. Helping people through breathing struggles is stressful under normal
conditions. For those therapists who work in hospitals, the stress is more constant. Administering
CPR and assisting with treatment in emergency rooms is high pressure.
Another factor that can add to the stress is working with children who have breathing issues, or
who have been in a drowning accident or other type of accident that has compromised their
respiration.
WORK CONDITION and ENVIRONMENT:
Nurses do believe that, RTs may have to work on weekends, holidays, and various shifts and
may also have to work extended hours in case of emergency. A career in respiratory therapy also
means being exposed to infections and micro-organisms.
Another misconception is that, as being a registered nurse, they do have more opportunities of
different places to work like hospitals, clinics, doctor\'s offices, jail\'s, factories, home health, and
of course as a school nurse. whereas, as RTs, well not so many as they could work on just
hospitals and home health.
These are the major misconceptions among nurses about a respiratory therapist.
Solution
A respiratory therapy professional diagnoses and treats those suffering from chronic and other
respiratory problems. Some of their patients with respiratory problems would be:
• Premature infants
• Asthma patients
• Patients with lung infections
• Heart patients
A respiratory therapist (RT) has a variety of .
a) Angular gyrus is present in parietal cortex. It is concerned with.pdf
a) Angular gyrus is present in parietal cortex. It is concerned with storage of memories of the
visual symbols.
b) Broca\'s aphasia occurs due to damage of left inferior frontal region of cerebral cortex.
Broca’s area:
It is area 44. Area 44 is the motor speech area. It is also called speech center. It is situated in the
lower part of prefrontal cortex.
Broca’s aphasia:
c) Wernicke\'s area is situated in upper part of temporal lobe. It plays an important role in
speech. Lesions in this area results in Wernicke\'s aphasia.
d) Damage to left temporal lobe leads to Wernicke\'s aphasia.
Therefore, option b is correct.
Broca’s aphasia is usually associated with the lesions of left inferior frontal region of cerebral
cortex.
Solution
a) Angular gyrus is present in parietal cortex. It is concerned with storage of memories of the
visual symbols.
b) Broca\'s aphasia occurs due to damage of left inferior frontal region of cerebral cortex.
Broca’s area:
It is area 44. Area 44 is the motor speech area. It is also called speech center. It is situated in the
lower part of prefrontal cortex.
Broca’s aphasia:
c) Wernicke\'s area is situated in upper part of temporal lobe. It plays an important role in
speech. Lesions in this area results in Wernicke\'s aphasia.
d) Damage to left temporal lobe leads to Wernicke\'s aphasia.
Therefore, option b is correct.
Broca’s aphasia is usually associated with the lesions of left inferior frontal region of cerebral
cortex..
Since youre diluting the solution to 2.5 times .pdf
Since you\'re diluting the solution to 2.5 times its initial volume, since VM =
constant, the final concentration is the initial concentration over 2.5, or .0250/2.5 = .010 M.
Solution
Since you\'re diluting the solution to 2.5 times its initial volume, since VM =
constant, the final concentration is the initial concentration over 2.5, or .0250/2.5 = .010 M..
wo (-2) above the O3 means there are 2 extra elec.pdf
wo (-2) above the O3 means there are 2 extra electrons available for bonding.
Polyatomic ions are groups of atoms joined together by covalent bonds, like molecules, except
that they have a charge. The only difference between the process for drawing Lewis diagrams for
molecules and for ions is in determining the number of available electrons. Count the valence as
before then add an electron for each unit of negative charge or subtract one for each unit of
positive charge. For example, the sulfite ion has the formula SO32-. The number of valence
electrons is 6 + 3*6 + 2 = 26. The atom with the highest electronegativity in the sulfite ion is the
sulfur, so it belongs in the center of the diagram with oxygen atoms bonded to it. Fill in enough
bonds to hold the ion together then add non-bonding electrons until all atoms obey the octet rule
or you have used all that are available.
Solution
wo (-2) above the O3 means there are 2 extra electrons available for bonding.
Polyatomic ions are groups of atoms joined together by covalent bonds, like molecules, except
that they have a charge. The only difference between the process for drawing Lewis diagrams for
molecules and for ions is in determining the number of available electrons. Count the valence as
before then add an electron for each unit of negative charge or subtract one for each unit of
positive charge. For example, the sulfite ion has the formula SO32-. The number of valence
electrons is 6 + 3*6 + 2 = 26. The atom with the highest electronegativity in the sulfite ion is the
sulfur, so it belongs in the center of the diagram with oxygen atoms bonded to it. Fill in enough
bonds to hold the ion together then add non-bonding electrons until all atoms obey the octet rule
or you have used all that are available..
More from Ankitagarwaleleraipu (20)
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