The most likely F1 will be the one that is made from 4 and 5. This i.pdf
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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.
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I dont understand numbers 14 and 15, please help.. PROBLEM BAS .pdf
I don\'t understand numbers 14 and 15, please help.. PROBLEM BAS 12, A plant of genotype is
testcrossed wil 18 If the two loci are 10 m apart, what proportion of progeny will be AB/ab 13.
The A locus and the Dlocus nked that no are so tightly recombination is ever observed between
if Ad 1. Ad is crossed with ab,aD d the F, is intercrossed, what phenotypes will be seen in the Fa
and in what proportions? 14. The R and Sloci are 35 m.u. apart. If a plant of genotype is selfed,
what progeny phenotypes will be seen and in what proportions 15. The cross EE. FF x efe fif s
de, and the F1 is then backcrossed with the recessive parent. The progeny gen. ocypes are
inferred from the phenotypes. The progeny genotypes, written as the gametic contributions of the
heterozygous parent, are in the following proportions: E. F 2 Explain these results. Progeny
Class phenotypes.
Solution
Answers:
14.
R S/r s * R S/r s
Gametes: 1/2 (1 – 0.35) R S
1/2 (1 – 0.35) r s
1/2 (0.35) R s
1/2 (0.35) r S
F1 genotypes will be:
0.1056 R S/R S
0.1138 r s/r S
0.1056 r s/r s
0.1138 r s/R s
0.2113 R S/r s
0.0306 R s/R s
0.1138 R S/r S
0.0306 r S/r S
0.1138 R S/R s
0.0613 R s/r S
F1 phenotypes will be:
0.6058 R S
0.1056 r s
0.1444 R s
0.1444 r S
15. The cross: E/e · F/f * e/e · f/f.
The progeny should be in a 1:1:1:1 ratio if independent assortment takes place, which is not
observed. So, linkage exists. E f and e F are recombinants and equals one-third of the progeny.
The two genes are 33.3 map units (m.u.) apart.
RF = 100% * 1/3 = 33.3%.
1. Meiosis And Dihybrid Cross
The document discusses meiosis, genetic crosses, and inheritance patterns. It begins by outlining meiosis and its role in genetic variation. It then discusses dihybrid crosses and Mendel's results, which showed independent assortment and ratios of 9:3:3:1. The document further explains linkage, recombination frequency, and genetic maps. It concludes by covering sex-linked inheritance and examples like color blindness and hemophilia.
Linkage and crossing over
This document discusses linkage and crossing over in genetics. It begins by outlining Mendel's laws of inheritance and then describes how Bateson and Punnett discovered linkage between genes for flower color and pollen shape in peas through deviations from expected Mendelian ratios. Morgan later showed with Drosophila experiments that linkage occurs when genes are located near each other on the same chromosome. The document then explains that crossing over during meiosis results in new combinations of maternal and paternal chromosomes, allowing genes to be reshuffled between generations. Linkage maps can be constructed by measuring the frequency of recombination between gene pairs, with less recombination indicating closer proximity on the chromosome.
Content
This document discusses apparent deviations from Mendel's laws of inheritance. It describes several genetic phenomena that can produce inheritance patterns different from classic Mendelian ratios, including polymery, complementary genes, epistasis, cryptomery, and gene interactions. Polymery involves multiple alleles at a single gene locus influencing a trait. Complementary genes require two dominant alleles together to express a phenotype. Epistasis occurs when one dominant gene masks the expression of another. Cryptomery means a trait is only expressed when two normally hidden dominant genes are present simultaneously. Gene interactions were demonstrated through experiments on chicken comb phenotypes.
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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
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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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I dont understand numbers 14 and 15, please help.. PROBLEM BAS .pdf
I don\'t understand numbers 14 and 15, please help.. PROBLEM BAS 12, A plant of genotype is
testcrossed wil 18 If the two loci are 10 m apart, what proportion of progeny will be AB/ab 13.
The A locus and the Dlocus nked that no are so tightly recombination is ever observed between
if Ad 1. Ad is crossed with ab,aD d the F, is intercrossed, what phenotypes will be seen in the Fa
and in what proportions? 14. The R and Sloci are 35 m.u. apart. If a plant of genotype is selfed,
what progeny phenotypes will be seen and in what proportions 15. The cross EE. FF x efe fif s
de, and the F1 is then backcrossed with the recessive parent. The progeny gen. ocypes are
inferred from the phenotypes. The progeny genotypes, written as the gametic contributions of the
heterozygous parent, are in the following proportions: E. F 2 Explain these results. Progeny
Class phenotypes.
Solution
Answers:
14.
R S/r s * R S/r s
Gametes: 1/2 (1 – 0.35) R S
1/2 (1 – 0.35) r s
1/2 (0.35) R s
1/2 (0.35) r S
F1 genotypes will be:
0.1056 R S/R S
0.1138 r s/r S
0.1056 r s/r s
0.1138 r s/R s
0.2113 R S/r s
0.0306 R s/R s
0.1138 R S/r S
0.0306 r S/r S
0.1138 R S/R s
0.0613 R s/r S
F1 phenotypes will be:
0.6058 R S
0.1056 r s
0.1444 R s
0.1444 r S
15. The cross: E/e · F/f * e/e · f/f.
The progeny should be in a 1:1:1:1 ratio if independent assortment takes place, which is not
observed. So, linkage exists. E f and e F are recombinants and equals one-third of the progeny.
The two genes are 33.3 map units (m.u.) apart.
RF = 100% * 1/3 = 33.3%.
1. Meiosis And Dihybrid Cross
The document discusses meiosis, genetic crosses, and inheritance patterns. It begins by outlining meiosis and its role in genetic variation. It then discusses dihybrid crosses and Mendel's results, which showed independent assortment and ratios of 9:3:3:1. The document further explains linkage, recombination frequency, and genetic maps. It concludes by covering sex-linked inheritance and examples like color blindness and hemophilia.
Linkage and crossing over
This document discusses linkage and crossing over in genetics. It begins by outlining Mendel's laws of inheritance and then describes how Bateson and Punnett discovered linkage between genes for flower color and pollen shape in peas through deviations from expected Mendelian ratios. Morgan later showed with Drosophila experiments that linkage occurs when genes are located near each other on the same chromosome. The document then explains that crossing over during meiosis results in new combinations of maternal and paternal chromosomes, allowing genes to be reshuffled between generations. Linkage maps can be constructed by measuring the frequency of recombination between gene pairs, with less recombination indicating closer proximity on the chromosome.
Content
This document discusses apparent deviations from Mendel's laws of inheritance. It describes several genetic phenomena that can produce inheritance patterns different from classic Mendelian ratios, including polymery, complementary genes, epistasis, cryptomery, and gene interactions. Polymery involves multiple alleles at a single gene locus influencing a trait. Complementary genes require two dominant alleles together to express a phenotype. Epistasis occurs when one dominant gene masks the expression of another. Cryptomery means a trait is only expressed when two normally hidden dominant genes are present simultaneously. Gene interactions were demonstrated through experiments on chicken comb phenotypes.
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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.
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2.
3.
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2.
3..
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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.
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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
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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
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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.
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..
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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..
Main Java[All of the Base Concepts}.docx
This is part 1 of my Java Learning Journey. This Contains Custom methods, classes, constructors, packages, multithreading , try- catch block, finally block and more.
Solutons Maths Escape Room Spatial .pptx
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5Solution5.pdf
The document discusses a solution but provides no details about the problem or type of solution. It only contains the number "5" repeated five times with the word "Solution" in the middle, so there is not enough contextual information to generate a meaningful multi-sentence summary.
1. The smart phone, computer, navigation system and digital scanner .pdf
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.
(1)Solution(1).pdf
This 1 sentence document does not provide enough context or information to generate a meaningful 3 sentence summary. The document consists of a single word - "Solution" - with no other details provided.
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.
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..
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What does absense of MHC Class I indicateAlmost all nucleated cel.pdf
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The probabilities must add up to one. Add the ones you have and su.pdf
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Subculturing or passage is the method of preserving or continuing th.pdf
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Marijuana is also called as cannabis and it is a preparation of cann.pdf
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IHD = 1 Therefore there can be only 1 -bond or ring. The two signals.pdf
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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.pdf
homedesigninsertanimationsdevelopmentSolutionhomedes.pdf
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B) The energy required to remove an electron from an atomSolutio.pdf
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AnswerThe following molecules would not require carrier transport.pdf
AnswerThe following molecules would not require carrier transport.pdf
AnswerCyanide is an irreversible enzyme inhibitor of Cytochrome C.pdf
AnswerCyanide is an irreversible enzyme inhibitor of Cytochrome C.pdf
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The webinar may also give some examples on how nonprofits can best leverage Walmart Business+.
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Walmart Business + (https://business.walmart.com/plus) is a new shopping experience for nonprofits, schools, and local business customers that connects an exclusive online shopping experience to stores. Benefits include free delivery and shipping, a 'Spend Analytics” feature, special discounts, deals and tax-exempt shopping.
Special TechSoup offer for a free 180 days membership, and up to $150 in discounts on eligible orders.
Spark Good (walmart.com/sparkgood) is a charitable platform that enables nonprofits to receive donations directly from customers and associates.
Answers about how you can do more with Walmart!"
How to Make a Field Mandatory in Odoo 17
In Odoo, making a field required can be done through both Python code and XML views. When you set the required attribute to True in Python code, it makes the field required across all views where it's used. Conversely, when you set the required attribute in XML views, it makes the field required only in the context of that particular view.
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
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তাই একজন নাগরিক হিসাবে এই তথ্য গুলো আপনার জানা প্রয়োজন ...।
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Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
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- 1. The most likely F1 will be the one that is made from 4 and 5. This is because see the map between c and d gene. In map 1 the distance is 15 and in map 2 the distance is 20. What we know is , the larger the distance between two genes there are more recombinants between those 2 genes. Here in map 1 the distance between c and d is 10 which means the 2 genes are more likely to be in a shorter dstance ie they are linked. Linked genes fail to separate during meiosis and results in lesser number of recombinants and more of parents. So here in 2x3 F1s we cannot say that the heterozygous will replace the homozygous recessive allele. But in map2 ie F1s of 4x5, the distance between c and d is 20, and there will be less linkage compared to F1s of 2x3 and hence there is a chance for deletion of homozygous recessive alleles in them. Solution The most likely F1 will be the one that is made from 4 and 5. This is because see the map between c and d gene. In map 1 the distance is 15 and in map 2 the distance is 20. What we know is , the larger the distance between two genes there are more recombinants between those 2 genes. Here in map 1 the distance between c and d is 10 which means the 2 genes are more likely to be in a shorter dstance ie they are linked. Linked genes fail to separate during meiosis and results in lesser number of recombinants and more of parents. So here in 2x3 F1s we cannot say that the heterozygous will replace the homozygous recessive allele. But in map2 ie F1s of 4x5, the distance between c and d is 20, and there will be less linkage compared to F1s of 2x3 and hence there is a chance for deletion of homozygous recessive alleles in them.