Drag each cell structure to the appropriate bin. If a structure is fo.pdf
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Drag each cell structure to the appropriate bin. If a structure is found in both plant cells and animal cells, drag it to the \"both\" bin. Solution Plant cell only Chloroplast, cellulose cell wall, Golgi apparatus, Animal cells only Centriole Both Mitochondria, Endoplasmic reticulum, nucleus, plasma membrane..
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Some Calvin Cycle enzymes contain disulfide bonds that must be reduc.pdf
Some Calvin Cycle enzymes contain disulfide bonds that must be reduced through a mechanism
involving thioredoxin in order for the enzyme to be active in sunlight. What is the name of the
key regulatory protein that inactivates these same Calvin Cycle enzymes by oxidation when the
sun goes down? Choose the ONE best answer.
Question 10 options:
Thioredoxin requires sunlight for proper folding into its functional three dimensional structure,
so in the absence of light, it is unfolded and cannot reduce these enzymes.
There is no regulatory protein that oxidizes these Calvin Cycle enzymes; oxidation is
spontaneous.
Sunlight is a reducing energy and moonlight is an oxidizing energy, so when the moon comes
out at night, the Calvin Cycle enzymes are oxidized spontaneously.
None of these answers are correct.
The protein is called ferredoxin-thioredoxin reductase.
Thioredoxin has multiple light-sensing properties and when the sun goes down it turns into a
strong oxidant.
Thioredoxin requires sunlight for proper folding into its functional three dimensional structure,
so in the absence of light, it is unfolded and cannot reduce these enzymes.
There is no regulatory protein that oxidizes these Calvin Cycle enzymes; oxidation is
spontaneous.
Sunlight is a reducing energy and moonlight is an oxidizing energy, so when the moon comes
out at night, the Calvin Cycle enzymes are oxidized spontaneously.
None of these answers are correct.
The protein is called ferredoxin-thioredoxin reductase.
Thioredoxin has multiple light-sensing properties and when the sun goes down it turns into a
strong oxidant.
Solution
Option 4 \" None of these answers are correct.\"
\"Chloroplast thioredoxin systems are crucial components of redox network, mediating
environmental signals to chloroplast proteins. One key protein in calvin cycle regulatory
processes is thioredoxin. The reduced form of thioredoxin activates many biosynthetic enzymes
by reducing disulfide bridges that control their activity and inhibits several degradative enzymes
by the same means. In chloroplasts, oxidized thioredoxin is reduced by ferredoxin in a reaction
catalyzed by ferredoxin-thioredoxin reductase. This enzyme contains a 4Fe-4S cluster that
couples two one-electron oxidations of reduced ferredoxin to the two-electron reduction of
thioredoxin. Thus, the activities of the light and dark reactions of photosynthesis are coordinated
through electron transfer from reduced ferredoxin to thioredoxin and then to component enzymes
containing regulatory disulfide bonds. Other means of control of calvin cycle also exist. For
instance, phosphoribulose kinase and glyceraldehyde 3-phosphate dehydrogenase also are
regulated by NADPH directly. Some plants synthesize a transition-state inhibitor,
carboxyarabinitol-1-phosphate (CA1P), in the dark. In the dark, these enzymes associate with a
small protein called CP12 to form a large complex in which the enzymes are inactivated. RuBP
Carboxylase Activase facilitates release.
Simplify and write the answer with positive exponents Simplify and w.pdf
Simplify and write the answer with positive exponents Simplify and write the answer with
positive exponents (4/5)^8(4/5)^-5(4/5)^-4
Solution
(4/5)^8(4/5)^-8(4/5)^-4
we can use the exponent property : a^x*a^y =a^(x+y)
So, (4/5)^8(4/5)^-8(4/5)^-4 = (4/5)^(8-5-4)
= (4/5)^(-1)
use another exponent property .So,
= (4/5)^(-1) = 5/4
=5/4.
Question 7 (of 20) 7. 1000 poants TB 1301 which of the following an.pdf
Question 7 (of 20) 7. 1000 poants TB 1301 which of the following analysis techniques d…
Which of the following analysis techniques does notexamines significant sustained changes over
time? O Horizontal analyses O Trend anayses O Vertical analyses O Time-series analyses
Mutiple Choice TB 13-01 Which of the following analyshs techniques d... yses to Recognize
Financial Changes that Unfold
Solution
Answer is Option (3) Vertical Analysis Verical analysis does not used to
examine the changes over time like horizental , trend and time series analysis. In the Veritical
analysis, we take every line item and then calculate the persentage over revenue. This process
helps to compare the financial statements of different companies.
Problem 12-9ACondensed financial data of Waterway Industries follo.pdf
Problem 12-9A
Condensed financial data of Waterway Industries follow.
Waterway Industries
Comparative Balance Sheets
December 31
Assets
2017
2016
$ 114,736
$ 68,728
124,676
53,960
159,750
146,047
40,328
36,920
195,960
154,780
404,700
344,350
(71,000
)
(73,840
)
$969,150
$730,945
Liabilities and Stockholders’ Equity
$ 144,840
$ 95,566
23,430
29,820
156,200
207,320
312,400
248,500
332,280
149,739
$969,150
$730,945
Waterway Industries
Income Statement Data
For the Year Ended December 31, 2017
$551,613
$192,353
17,622
66,030
38,738
6,717
10,650
332,110
$ 219,503
Additional information:
Prepare a statement of cash flows using the indirect method. (Show amounts that decrease cash
flow with either a - sign e.g. -15,000 or in parenthesis e.g. (15,000).)
Adjustments to reconcile net income to
Waterway Industries
Comparative Balance Sheets
December 31
Assets
2017
2016Cash
$ 114,736
$ 68,728Accounts receivable
124,676
53,960Inventory
159,750
146,047Prepaid expenses
40,328
36,920Long-term investments
195,960
154,780Plant assets
404,700
344,350Accumulated depreciation
(71,000
)
(73,840
)Total
$969,150
$730,945
Liabilities and Stockholders’ EquityAccounts payable
$ 144,840
$ 95,566Accrued expenses payable
23,430
29,820Bonds payable
156,200
207,320Common stock
312,400
248,500Retained earnings
332,280
149,739Total
$969,150
$730,945
Solution
Waterway Industries Statement of Cash Flows For the year ended Amount Cash
flows from operating activities Net Income $219,503 Adjust to reconcile net income to net
cash provided Depreciation expenses $66,030 Loss on disposal of plant assets $10,650
Increase in Accounts receivable -$70,716 Increase in Inventory -$13,703 Increase in
Prepaid expenses -$3,408 Increase in Accounts Payable $49,274 Decrease in accrued
expenses payable -$6,390 Net cash used by operating activities $251,240 Cash flows
from investing activities Cash payment for acquisition of plant assets -$142,000 Sale of
Plant Assets $2,130 Cash payment for acquisition of Long-term investment -$41,180 Net
cash used by investing activities -$181,050 Cash flows from financing activities
Issuance of Common Stock $63,900 Bond Payable matured -$51,120 Payment of Dividends
-$36,962 Net cash provided by financing activities -$24,182 Net decrease in cash
$46,008 Cash at Beginning of Period $68,728 Cash at End of Period $114,736.
Please help me with these 3 questions! True or False. Assortment of .pdf
Please help me with these 3 questions! True or False. Assortment of chromosomes and
crossovers together generate new combinations of alleles that differ from the parental
combinations. True False True or False. If two genes are very far apart on the same
chromosome, they can sometimes produce as many recombinants as genes that reside on
different chromosomes. True False Mendel performed a monohybrid cross by crossing two
pure-breeding parental lines, and then crossing the F1 individuals to produce an F2. He predicted
that the F2 would contain a mix of heterozygous (Aa) and homozygous dominant (AA)
individuals. How might his hypothesis be tested? Use the F2 offspring as one parent in a
dihybrid cross Perform testcrosses with each of several of the dominant phenotype F2 offspring
crossed to a homozygous recessive genotype Use F2 offspring as one parent in a trihybrid cross
Perform testcrosses with each of several of the dominant phenotype F2 offspring crossed to a
homozygous dominant genotype None of the above.
Solution
40.
The given statement is true. Independent assortment law is proposed by Mendel states that the
different pairs of alleles are passed to offspring independently of each other resulting in new
combinations of genes in the offspring that are different from parents. Crossing over is another
phenomenon in which homologous chromosomes exchange segments resulting in the formation
of new combination of genes in offspring and are different from parents.
41.
The given statement is true. The genes undergoing independent assortment may produce 50%
recombinants whether the genes are located on nonhomologous chromosomes or genes are
located far apart in a single chromosome.
42.
The result of F1 cross is expected having F2 individuals of both heterozygous and homozygous
dominant individuals is tested by performing test cross between dominant phenotype in F2
offspring with a homozygous recessive genotype. Option 2 is correct..
Performance BudgetingPerformance budgeting has been attempted at t.pdf
Performance Budgeting
Performance budgeting has been attempted at the local level in recent years. Address the issues
of performance budgeting while answering the following questions: What attributes of
performance budgeting make it particularly suitable to local government budgeting? Will the
same attributes be as useful at the federal level?
Solution
Performance Budgetingmeasures the activities in terms of output. It shows cost matching with
operation and the focus on decision making and evaluation is on output rather than merely on
expenditure. Performance Budgeting provide a meaningful relationship between estimated inputs
and expected outputs as an integral part of the budgeting system.
Attributes of performance budgeting which make it particularly suitable to local government
budgeting are as follows:-
1) Performance budgeting focuses at establishing a relationship between the inputs and the
outputs.
2) Traditional budgets highlights only the items of expenditure, namely salaries, stores,
materials, rates, rent etc. Whereas Performance budgeting emphasises more on the functions of
organisation, the programmes to discharge these functions and the activities which will be
involved in undertaking these programmes.
3) Performance budgeting facilitate to provide information on what the government is doing and
what it gets for the money spent.
4) Performance budgeting is related to management efficiency especially in the government
affairs.
These all attributes are useful at the federal level..
Many bridges are made from rebar-reinforced concrete composites. Cla.pdf
Many bridges are made from rebar-reinforced concrete composites. Classify the composite
system (fiber-reinforced, particulate, laminar, or hybrid), and explain why it is such a ubiquitous
choice of construction.
Solution
Fibers or particles embedded in matrix of another material are the best example of modern-day
composite materials, which are mostly structural. Laminates are composite material where
different layers of materials give them the specific character of a composite material having a
specific function to perform. Fabrics have no matrix to fall back on, but in them, fibers of
different compositions combine to give them a specific character. Reinforcing materials
generally withstand maximum load and serve the desirable properties. Further, though composite
types are often distinguishable from one another, no clear determination can be really made. To
facilitate definition, the accent is often shifted to the levels at which differentiation take place
viz., microscopic or macroscopic. In matrix-based structural composites, the matrix serves two
paramount purposes viz., binding the reinforcement phases in place and deforming to distribute
the stresses among the constituent reinforcement materials under an applied force.
Composite materials are commonly classified at following two distinct levels:
• The first level of classification is usually made with respect to the matrix constituent. The
major composite classes include Organic Matrix Composites (OMCs), Metal Matrix Composites
(MMCs) and Ceramic Matrix Composites (CMCs). The term organic matrix composite is
generally assumed to include two classes of composites, namely Polymer Matrix Composites
(PMCs) and carbon matrix composites commonly referred to as carboncarbon composites.
• The second level of classification refers to the reinforcement form - fibre reinforced
composites, laminar composites and particulate composites. Fibre Reinforced composites (FRP)
can be further divided into those containing discontinuous or continuous fibres.
The role of the reinforcement in a composite material
• Fibre Reinforced Composites are composed of fibres embedded in matrix material. Such a
composite is considered to be a discontinuous fibre or short fibre composite if its properties vary
with fibre length. On the other hand, when the length of the fibre is such that any further increase
in length does not further increase, the elastic modulus of the composite, the composite is
considered to be continuous fibre reinforced. Fibres are small in diameter and when pushed
axially, they bend easily although they have very good tensile properties. These fibres must be
supported to keep individual fibres from bending and buckling.
• Laminar Composites are composed of layers of materials held together by matrix. Sandwich
structures fall under this category.
• Particulate Composites are composed of particles distributed or embedded in a matrix body.
The particles may be flakes or in powder form. Concrete and .
If the cystic fibrosis allele protects against tuberculosis the same .pdf
If the cystic fibrosis allele protects against tuberculosis the same way the sickle-cell allele
protects against malaria, then which of the following should be true of a comparison between
regions with and without tuberculosis? Cystic fibrosis deaths should be more common in regions
with tuberculosis. Cystic fibrosis deaths should be less common in regions with tuberculosis.
Cystic fibrosis deaths should be equally common in both types of regions. Regional differences
in the cystic fibrosis death rate should be purely random and unpredictable. A person who is
heterozygous for the cystic fibrosis allele moves to a small, isolated community where no one
previously carried the allele. If the cystic fibrosis allele protects against tuberculosis the same
way the sickle-cell allele protects against malaria, what should happen to the frequency of the
cystic fibrosis allele in the community over time, and why? the cystic fibrosis allele should
disappear from the population, because a single individual with the allele is not enough for it to
proliferate. the cystic fibrosis allele should increase to a relatively high frequency, because
heterozygotes with the allele will be more likely to survive than others. the cystic fibrosis allele
should become fixed in the population, due to genetic drift. the cystic fibrosis allele should either
disappear or increase in frequency, depending on chance as well as on tuberculosis prevalence
and death rate.
Solution
5.8 Cystic fibrosis is a autosomal recessive disease. Cystic fibrosis allele protects against
tuberculosis similar to sickle cell allele protects against Malaria. The cystic fibrosis death rate
cannot be predicted in regions with tuberculosis and regional differences influencing cystic
fibrosis death rate are random. This is because the cystic fibrosis allel frequency depends o
prevalence of tuberculosis. Option 4 is correct.
5.9 The cystic fibrosis allele frequency depends on prevalence of tuberculosis in the given
community. The allele frequency either may increase due to prevalence of tuberculosis or the
allele causing cystic fibrosis may disappear because of frequency of dominant alleles in the
community. Therefore, option 4 is correct..
Recommended
Some Calvin Cycle enzymes contain disulfide bonds that must be reduc.pdf
Some Calvin Cycle enzymes contain disulfide bonds that must be reduced through a mechanism
involving thioredoxin in order for the enzyme to be active in sunlight. What is the name of the
key regulatory protein that inactivates these same Calvin Cycle enzymes by oxidation when the
sun goes down? Choose the ONE best answer.
Question 10 options:
Thioredoxin requires sunlight for proper folding into its functional three dimensional structure,
so in the absence of light, it is unfolded and cannot reduce these enzymes.
There is no regulatory protein that oxidizes these Calvin Cycle enzymes; oxidation is
spontaneous.
Sunlight is a reducing energy and moonlight is an oxidizing energy, so when the moon comes
out at night, the Calvin Cycle enzymes are oxidized spontaneously.
None of these answers are correct.
The protein is called ferredoxin-thioredoxin reductase.
Thioredoxin has multiple light-sensing properties and when the sun goes down it turns into a
strong oxidant.
Thioredoxin requires sunlight for proper folding into its functional three dimensional structure,
so in the absence of light, it is unfolded and cannot reduce these enzymes.
There is no regulatory protein that oxidizes these Calvin Cycle enzymes; oxidation is
spontaneous.
Sunlight is a reducing energy and moonlight is an oxidizing energy, so when the moon comes
out at night, the Calvin Cycle enzymes are oxidized spontaneously.
None of these answers are correct.
The protein is called ferredoxin-thioredoxin reductase.
Thioredoxin has multiple light-sensing properties and when the sun goes down it turns into a
strong oxidant.
Solution
Option 4 \" None of these answers are correct.\"
\"Chloroplast thioredoxin systems are crucial components of redox network, mediating
environmental signals to chloroplast proteins. One key protein in calvin cycle regulatory
processes is thioredoxin. The reduced form of thioredoxin activates many biosynthetic enzymes
by reducing disulfide bridges that control their activity and inhibits several degradative enzymes
by the same means. In chloroplasts, oxidized thioredoxin is reduced by ferredoxin in a reaction
catalyzed by ferredoxin-thioredoxin reductase. This enzyme contains a 4Fe-4S cluster that
couples two one-electron oxidations of reduced ferredoxin to the two-electron reduction of
thioredoxin. Thus, the activities of the light and dark reactions of photosynthesis are coordinated
through electron transfer from reduced ferredoxin to thioredoxin and then to component enzymes
containing regulatory disulfide bonds. Other means of control of calvin cycle also exist. For
instance, phosphoribulose kinase and glyceraldehyde 3-phosphate dehydrogenase also are
regulated by NADPH directly. Some plants synthesize a transition-state inhibitor,
carboxyarabinitol-1-phosphate (CA1P), in the dark. In the dark, these enzymes associate with a
small protein called CP12 to form a large complex in which the enzymes are inactivated. RuBP
Carboxylase Activase facilitates release.
Simplify and write the answer with positive exponents Simplify and w.pdf
Simplify and write the answer with positive exponents Simplify and write the answer with
positive exponents (4/5)^8(4/5)^-5(4/5)^-4
Solution
(4/5)^8(4/5)^-8(4/5)^-4
we can use the exponent property : a^x*a^y =a^(x+y)
So, (4/5)^8(4/5)^-8(4/5)^-4 = (4/5)^(8-5-4)
= (4/5)^(-1)
use another exponent property .So,
= (4/5)^(-1) = 5/4
=5/4.
Question 7 (of 20) 7. 1000 poants TB 1301 which of the following an.pdf
Question 7 (of 20) 7. 1000 poants TB 1301 which of the following analysis techniques d…
Which of the following analysis techniques does notexamines significant sustained changes over
time? O Horizontal analyses O Trend anayses O Vertical analyses O Time-series analyses
Mutiple Choice TB 13-01 Which of the following analyshs techniques d... yses to Recognize
Financial Changes that Unfold
Solution
Answer is Option (3) Vertical Analysis Verical analysis does not used to
examine the changes over time like horizental , trend and time series analysis. In the Veritical
analysis, we take every line item and then calculate the persentage over revenue. This process
helps to compare the financial statements of different companies.
Problem 12-9ACondensed financial data of Waterway Industries follo.pdf
Problem 12-9A
Condensed financial data of Waterway Industries follow.
Waterway Industries
Comparative Balance Sheets
December 31
Assets
2017
2016
$ 114,736
$ 68,728
124,676
53,960
159,750
146,047
40,328
36,920
195,960
154,780
404,700
344,350
(71,000
)
(73,840
)
$969,150
$730,945
Liabilities and Stockholders’ Equity
$ 144,840
$ 95,566
23,430
29,820
156,200
207,320
312,400
248,500
332,280
149,739
$969,150
$730,945
Waterway Industries
Income Statement Data
For the Year Ended December 31, 2017
$551,613
$192,353
17,622
66,030
38,738
6,717
10,650
332,110
$ 219,503
Additional information:
Prepare a statement of cash flows using the indirect method. (Show amounts that decrease cash
flow with either a - sign e.g. -15,000 or in parenthesis e.g. (15,000).)
Adjustments to reconcile net income to
Waterway Industries
Comparative Balance Sheets
December 31
Assets
2017
2016Cash
$ 114,736
$ 68,728Accounts receivable
124,676
53,960Inventory
159,750
146,047Prepaid expenses
40,328
36,920Long-term investments
195,960
154,780Plant assets
404,700
344,350Accumulated depreciation
(71,000
)
(73,840
)Total
$969,150
$730,945
Liabilities and Stockholders’ EquityAccounts payable
$ 144,840
$ 95,566Accrued expenses payable
23,430
29,820Bonds payable
156,200
207,320Common stock
312,400
248,500Retained earnings
332,280
149,739Total
$969,150
$730,945
Solution
Waterway Industries Statement of Cash Flows For the year ended Amount Cash
flows from operating activities Net Income $219,503 Adjust to reconcile net income to net
cash provided Depreciation expenses $66,030 Loss on disposal of plant assets $10,650
Increase in Accounts receivable -$70,716 Increase in Inventory -$13,703 Increase in
Prepaid expenses -$3,408 Increase in Accounts Payable $49,274 Decrease in accrued
expenses payable -$6,390 Net cash used by operating activities $251,240 Cash flows
from investing activities Cash payment for acquisition of plant assets -$142,000 Sale of
Plant Assets $2,130 Cash payment for acquisition of Long-term investment -$41,180 Net
cash used by investing activities -$181,050 Cash flows from financing activities
Issuance of Common Stock $63,900 Bond Payable matured -$51,120 Payment of Dividends
-$36,962 Net cash provided by financing activities -$24,182 Net decrease in cash
$46,008 Cash at Beginning of Period $68,728 Cash at End of Period $114,736.
Please help me with these 3 questions! True or False. Assortment of .pdf
Please help me with these 3 questions! True or False. Assortment of chromosomes and
crossovers together generate new combinations of alleles that differ from the parental
combinations. True False True or False. If two genes are very far apart on the same
chromosome, they can sometimes produce as many recombinants as genes that reside on
different chromosomes. True False Mendel performed a monohybrid cross by crossing two
pure-breeding parental lines, and then crossing the F1 individuals to produce an F2. He predicted
that the F2 would contain a mix of heterozygous (Aa) and homozygous dominant (AA)
individuals. How might his hypothesis be tested? Use the F2 offspring as one parent in a
dihybrid cross Perform testcrosses with each of several of the dominant phenotype F2 offspring
crossed to a homozygous recessive genotype Use F2 offspring as one parent in a trihybrid cross
Perform testcrosses with each of several of the dominant phenotype F2 offspring crossed to a
homozygous dominant genotype None of the above.
Solution
40.
The given statement is true. Independent assortment law is proposed by Mendel states that the
different pairs of alleles are passed to offspring independently of each other resulting in new
combinations of genes in the offspring that are different from parents. Crossing over is another
phenomenon in which homologous chromosomes exchange segments resulting in the formation
of new combination of genes in offspring and are different from parents.
41.
The given statement is true. The genes undergoing independent assortment may produce 50%
recombinants whether the genes are located on nonhomologous chromosomes or genes are
located far apart in a single chromosome.
42.
The result of F1 cross is expected having F2 individuals of both heterozygous and homozygous
dominant individuals is tested by performing test cross between dominant phenotype in F2
offspring with a homozygous recessive genotype. Option 2 is correct..
Performance BudgetingPerformance budgeting has been attempted at t.pdf
Performance Budgeting
Performance budgeting has been attempted at the local level in recent years. Address the issues
of performance budgeting while answering the following questions: What attributes of
performance budgeting make it particularly suitable to local government budgeting? Will the
same attributes be as useful at the federal level?
Solution
Performance Budgetingmeasures the activities in terms of output. It shows cost matching with
operation and the focus on decision making and evaluation is on output rather than merely on
expenditure. Performance Budgeting provide a meaningful relationship between estimated inputs
and expected outputs as an integral part of the budgeting system.
Attributes of performance budgeting which make it particularly suitable to local government
budgeting are as follows:-
1) Performance budgeting focuses at establishing a relationship between the inputs and the
outputs.
2) Traditional budgets highlights only the items of expenditure, namely salaries, stores,
materials, rates, rent etc. Whereas Performance budgeting emphasises more on the functions of
organisation, the programmes to discharge these functions and the activities which will be
involved in undertaking these programmes.
3) Performance budgeting facilitate to provide information on what the government is doing and
what it gets for the money spent.
4) Performance budgeting is related to management efficiency especially in the government
affairs.
These all attributes are useful at the federal level..
Many bridges are made from rebar-reinforced concrete composites. Cla.pdf
Many bridges are made from rebar-reinforced concrete composites. Classify the composite
system (fiber-reinforced, particulate, laminar, or hybrid), and explain why it is such a ubiquitous
choice of construction.
Solution
Fibers or particles embedded in matrix of another material are the best example of modern-day
composite materials, which are mostly structural. Laminates are composite material where
different layers of materials give them the specific character of a composite material having a
specific function to perform. Fabrics have no matrix to fall back on, but in them, fibers of
different compositions combine to give them a specific character. Reinforcing materials
generally withstand maximum load and serve the desirable properties. Further, though composite
types are often distinguishable from one another, no clear determination can be really made. To
facilitate definition, the accent is often shifted to the levels at which differentiation take place
viz., microscopic or macroscopic. In matrix-based structural composites, the matrix serves two
paramount purposes viz., binding the reinforcement phases in place and deforming to distribute
the stresses among the constituent reinforcement materials under an applied force.
Composite materials are commonly classified at following two distinct levels:
• The first level of classification is usually made with respect to the matrix constituent. The
major composite classes include Organic Matrix Composites (OMCs), Metal Matrix Composites
(MMCs) and Ceramic Matrix Composites (CMCs). The term organic matrix composite is
generally assumed to include two classes of composites, namely Polymer Matrix Composites
(PMCs) and carbon matrix composites commonly referred to as carboncarbon composites.
• The second level of classification refers to the reinforcement form - fibre reinforced
composites, laminar composites and particulate composites. Fibre Reinforced composites (FRP)
can be further divided into those containing discontinuous or continuous fibres.
The role of the reinforcement in a composite material
• Fibre Reinforced Composites are composed of fibres embedded in matrix material. Such a
composite is considered to be a discontinuous fibre or short fibre composite if its properties vary
with fibre length. On the other hand, when the length of the fibre is such that any further increase
in length does not further increase, the elastic modulus of the composite, the composite is
considered to be continuous fibre reinforced. Fibres are small in diameter and when pushed
axially, they bend easily although they have very good tensile properties. These fibres must be
supported to keep individual fibres from bending and buckling.
• Laminar Composites are composed of layers of materials held together by matrix. Sandwich
structures fall under this category.
• Particulate Composites are composed of particles distributed or embedded in a matrix body.
The particles may be flakes or in powder form. Concrete and .
If the cystic fibrosis allele protects against tuberculosis the same .pdf
If the cystic fibrosis allele protects against tuberculosis the same way the sickle-cell allele
protects against malaria, then which of the following should be true of a comparison between
regions with and without tuberculosis? Cystic fibrosis deaths should be more common in regions
with tuberculosis. Cystic fibrosis deaths should be less common in regions with tuberculosis.
Cystic fibrosis deaths should be equally common in both types of regions. Regional differences
in the cystic fibrosis death rate should be purely random and unpredictable. A person who is
heterozygous for the cystic fibrosis allele moves to a small, isolated community where no one
previously carried the allele. If the cystic fibrosis allele protects against tuberculosis the same
way the sickle-cell allele protects against malaria, what should happen to the frequency of the
cystic fibrosis allele in the community over time, and why? the cystic fibrosis allele should
disappear from the population, because a single individual with the allele is not enough for it to
proliferate. the cystic fibrosis allele should increase to a relatively high frequency, because
heterozygotes with the allele will be more likely to survive than others. the cystic fibrosis allele
should become fixed in the population, due to genetic drift. the cystic fibrosis allele should either
disappear or increase in frequency, depending on chance as well as on tuberculosis prevalence
and death rate.
Solution
5.8 Cystic fibrosis is a autosomal recessive disease. Cystic fibrosis allele protects against
tuberculosis similar to sickle cell allele protects against Malaria. The cystic fibrosis death rate
cannot be predicted in regions with tuberculosis and regional differences influencing cystic
fibrosis death rate are random. This is because the cystic fibrosis allel frequency depends o
prevalence of tuberculosis. Option 4 is correct.
5.9 The cystic fibrosis allele frequency depends on prevalence of tuberculosis in the given
community. The allele frequency either may increase due to prevalence of tuberculosis or the
allele causing cystic fibrosis may disappear because of frequency of dominant alleles in the
community. Therefore, option 4 is correct..
Match the graph of the sine function to the equation over [-4 Pi. 4 P.pdf
Match the graph of the sine function to the equation over [-4 Pi. 4 Pi]
Solution
here vertical shift is 1/2
from the graph
Amplitude = 3/2
period = =pi
phase shift is = -pi/4/pi= -1/4
so we can write Y = Asin(BX+C) + D
where A= amplitude
period = 2pi/B
phase shift = -C/B
vertical shift (D)=1/2
so y = 3/2sin(1/2 x -pi/4) +1/2
so it is option A
you can find answer in other way also
from the graph take some points
option A is the correct answer
plug x =-pi/4 then we get y=-1 and in the graph also y=-1.
location in the stem, function, and Chemical Composition of their Wa.pdf
location in the stem, function, and Chemical Composition of their Wall:
-Epidermis
-Periderm
-Xylem
- Phloem
- Parenchyma
- Collenchyma
- Sclerenchyma
Solution
Epidermis-
It is the single layer cell covering which makes the dermal tissue and covers the stem. It is the
outermost layer of the primary plant body.
Function- regulates exchange of gases in plants, prevents the loss of water, and helps in secreting
various metabolic compounds.
Chemical composition- hydrocarbon polymer impregnated with wax and lipids
Periderm-
Periderm is the corky outer layer of the stem. It is the secondary protective dermal tissue which
replaces epidermis during the growth thickness of the stem. Periderm is the multilayer tissue
system.
Function- Protects the stem from pathogen attack an dessication
Chemical composition- polysaccharides, tannins, biopolymers
Xylem-
Xylem is a type of transport tissue located in the vascular bundles.
Function-main function of xylem is to transport water from roots to shoots and leaves along with
some important nutrients.
Chemical composition- hemicelluloses, ligin, cellulose, pectin
Phloem-
Phloem is the living tissue found in the vascular bundles of the plants. It is the innermost layer of
the bark.
Function- the important function is to transport organic compounds formed during the process of
photosynthesis..
If Zn Binomial (n, ) , and n~possion(), find the distribution of Z .pdf
If Z|n Binomial (n, ) , and n~possion(), find the distribution of Z .
Is it more or less dispersed than the distribution of Z|n ?
Solution
since we have a binomial distribution with n and pi
and n has a poisson distribution
Z will have a normal distribution with mean = n by landha
it is more dispersed bvecause n depends of landha.
Given the global financial crisis of 2007–2009, do you anticipate an.pdf
Given the global financial crisis of 2007–2009, do you anticipate any changes to the systems of
fixed exchange rates and forward contracts in the near future? Are the changes you envision
strictly procedural or regulatory, or do you believe that some of these changes will intend to
provide safeguards against ethical lapses or loopholes?
Solution
As a result of the global financial crisis of 2007-2009, it showed that the protection companies
and investors have at the time to hedge against exchange rates instability were severely lacking
resulting in billions of dollars in lost wealth globally. Any financial market carries risks for its
users, but the main function of some markets is redistribution of risk. This is exactly what
forward markets and derivative markets do. They are more risky, but at the same time it is where
risks may be reduced or eliminated. Since the emergence of derivatives market, hedging
operations are increasingly performed in the futures and forward markets (due to minor
transaction costs, more expedient transactions, etc). Investors use forwards with a view to
protecting a certain yield level through the transfer of risk to other trade participants. This risk is
undertaken by speculators who take opposite positions from the hedgers and thus make the
market liquid. Hedgers, however, protect themselves by the transfer of risk to other participants
(speculators) against losses, but also reduce the possibility of profit increase.
Financial crises have many common features. In the background real economy, there is usually
the presence of an asset price “bubble” (or asset price inflation, for purists), a corresponding
credit boom, and large capital inflows into that economy (see, for example, Reinhart and Rogoff,
2008). However, these characteristics are necessary, but not sufficient, for a financial crisis to
develop. The severity of the crisis depends crucially on the underlying financial sector’s
exposure to these conditions and, in fact, the overall market’s uncertainty about the financial
sector’s exposure to them. A key role of financial regulation is to put limits on financial
institutions, so as to limit this exposure. While there are many reasons for the relative calm of the
U.S. financial system during the 50 years after the Great Depression, many analysts continue to
give credit to the financial regulation that was enacted at that time.
Given their inherently high leverage and the ease with which the risk-profile of financial assets
can be altered, banks and financial institutions have incentives to take on excessive risks.
Ordinarily, market mechanisms would be expected to price risks correctly and thereby ensure
that risk-taking in the economy is at efficient levels. However, there are two factors that have
impeded such efficient outcomes. First, with the repeal of most protections from the Banking Act
of the 1933, the only remaining protection against risk-shifting is capital requirements. If the
guarantees are m.
Generativity versus stagnation Early adult transitio.pdf
Generativity versus stagnation
Early adult transition
Intimacy versus isolation
Self-awareness transition
Generativity involves a concern for and interest in establishing and guiding the next
generation.
Generativity involves a commitment to improve the life conditions of future generations.
Generativity involves a commitment towards protecting and enhancing the conditions of
one’s society.
All of the above are true.
Socializing versus sexualizing in human relationships
Emotional flexibility versus emotional impoverishment
Culminating life structure versus early adult transition
Valuing wisdom versus valuing physical powers
Sexualizing in human relationships
Valuing physical powers
Emotional impoverishment
Mental flexibility
Early adult transition
Entry life structure for early adulthood
Culminating life structure for early adulthood
Entry life structure for middle adulthoodErikson theorized that after young people develop a
sense of identity, they next face the psychosocial crisis of: (Points : 2)
Solution
Soln:
1) Intimacy versus isolation
2) All of the above are true.
3) Culminating life structure versus early adult transition
4) Mental flexibility
5) Early adult transition
6) Entry life structure for middle adulthood.
Genetics Question 2 In the insect sex determination system, th.pdf
Genetics
Question 2 In the insect sex determination system, the are considered the homogametic sex
because they have 2X chromosomes females males
Solution
In insects, mostly the sex determination is of XX-XY type. In this, females are homogametic i.e.
they contain two copies of the same kind of sex chromosomes. Females contain two X
chromosomes. Males are heterogametic i.e. they contain two different kinds of sex
chromosomes. Hence, males contain one X chromosome and one Y chromosome.
Therefore, the correct answer is females..
Explain why malaria infection may lead to anemia. How would you clas.pdf
Explain why malaria infection may lead to anemia. How would you classify the anemia that
results?(scientific name for the type of anemia that results)
Solution
Plasmodium falciperum and Plasmodium vivax causes malaria in human and also responsible for
anemia .It is due to increase removal of erythrocyte as well as decreased production of
erythrocytes in the the bone marrow.Anemia is due to the continous removal/ destruction of
erythrocytes in the bone marrow. Parasitie infect RBC and its toxin causes harmful effect to the
erythrocyte. It causes severe anemia in human..
Figure 13.2 of a single pair of homologous chromosomes as they might.pdf
Figure 13.2 of a single pair of homologous chromosomes as they might appear during various
stages of either mitosis or meiosis 53) Which diagram(s) represent anaphase II of meiosis? A) II
only B) III only C) IV only D) V only E) either II or V Answer: D What\'s wrong with E?? If III
V VI /1 /N
Solution
It Showing various stages of Meiosis because the stages showing 2 anaphases, crossing over and
chiasma formation
Yaa the answer is D. Anaphase 2 occur in haploid cells, so it contain single chromosome but 2
chromatids. The two chromatids seperate in the Anaphase II. So D is correct.
Every individual produced by sexual reproduction is unique. This is .pdf
Every individual produced by sexual reproduction is unique. This is because of… (circle the
TWO correct answers)
a) Independent assortment of chromosomes during meiosis.
b) Independent assortment of chromosomes during mitosis.
c) Recombination occurring in prophase.
d) Separation of chromatids occurring in anaphase 2.
Solution
a. Independent assortment of chromosomes during Meiosis
c. Recombination occuring in prophase.
After sexual reproduction the gametes that unite to form zygote are haploid. Since the haploid
gametes are formed by the reductional divisions of diploid cell, So, The chromosomes that
haploid gamete will carry, will be responsible for the characters inherited to the progeny
Also the recombination that takes place in the prophase, determines the characters or traits
inherited after the crossing over of chromosomes.
Thus these two factors will be most responsible in determining the individuality of the progeny..
Explain to them the probability of future offspring being normal or h.pdf
Explain to them the probability of future offspring being normal or having the disease (assume
that this is an autosomal disease). Be sure to tell them whether the disease is brought about by a
dominant or recessive allele and what fraction of their children (if any) could be carriers for the
disease. Be sure to incorporate the following into your explanation: DNA and chromosomes,
genes, dominant and recessive alleles, paired alleles on homologous chromosomes, and
segregation of alleles during gamete production. You can use words and drawings/diagrams.
Sally and Steve are concerned about cystic fibrosis. Neither of them has it, nor do their parents or
grandparents, yet each one has a sibling with cystic fibrosis (Sally\'s sister and Steve\'s brother).
Sally and Steve want to have a second child (the first one, age 4, appears unaffected). Jess and
Jessie want to get married and have children. Unfortunately, they have the same great-
grandmother (not the same great-grandfather). The great-grandmother suffered from polycystic
kidney disease. No other family members suffered from the disease, but they were told
(correctly) that their great-grandmother had passed on the possibility of the disease.
Solution
Ques-1:
Aetiology and Pathophysiology: It is autosomal recessive disease (mode of inheritance), which is
familial. Gene for cystic fibrosis is located on the chromosome 7 & this recessive allele
propogated to the offspring via parental gamete formation by reduction division & problem will
be at segregation of chromosomes. A protein called, CF transmembrane -this gene produces
regulator (CFTR). In case of Pleiotrophy one affects 2 or more apparently not linked phenotypic
traits there by more gene could be affected (cystic fibrosis).
For instance, Sally and Steve are concerned about cystic fibrosis. Neither of them has it, nor do
their parents or grandparents, yet each one has a sibling with cystic fibrosis (Sally\'s sister and
Steve\'s brother) i.e. both Sally & Steve are carriers of the disease. Therefore, 2/3 (67%), of
probability with unaffected child. Sally and Steve want to have a second child but the first child
age-4 has no cystic fibrosis
Cystic fibrosis is an autosomal recessive condition. In the given example, the unaffected parents
will have an affected children Assume that the gene coding for the normal allele is C, and the
gene coding for the cystic fibrosis is c and C is dominant over c. Now, cross between two
heterozygotes will have the offspring with the following genotypes.
Cc* Cc = CC, Cc, Cc, cc (means, 3/4th are unaffected (out of these 2/4 are carriers) and the ¼
are affected -having disease).
C
c
C
CC
Cc
c
Cc
cc
C
c
C
CC
Cc
c
Cc
cc.
explain conceptual questionsvariabilitystandard devationz-sc.pdf
explain conceptual questions
variability?
standard devation?
z-scores (include a detailed drawing of a sample z-distribution)?
what is factorial design and why do we use them in psychology?
hypothesis testing( explain the null and alternative hypotheses- which one do we test against in
order to determin hte siginficance, alpha levels, and how probability is used to determin
significance)?
Solution.
Explain the difference between fermentation and respiration in biolo.pdf
Explain the difference between fermentation and respiration in biology?
Solution
Fermentation and Respiration are the two widely used processes in biology.
Respiration : It is a biochemical process in which food molecules are broken down by cells of an
organism with the release of carbon dioxide , water and ATP ( energy currency of cells).
Food + oxygen gives carbondioxide + water + ATP.
Types : 1. Aerobic respiration : Occurs in presence of oxygen, performed by eukaryotic cells,
yields 38 ATP molecules and Occurs in both cytosol and mitochondria.
2 : Anaerobic Respiration : Occurs in absence of oxygen, performed by prokaryotic cells, yields
only 2 ATP molecules and Occurs in cytosol only.
Fermentation : Term ist used by Louis Pasteur in 19th century.
It is a metabolic process in which sugar is converted to acids or alcohol.
It occurs in yeast and bacteria and sometimes in muscle cells ( lactic acid fermentation )
Fermentation Occurs when electron transport chain becomes defunct due to lack of final electron
acceptor I,e oxygen.
Difference between Respiration and Fermentation:
1. Respiration uses oxygen as electron acceptor to form ATP while fermentation uses inorganic
donars ( Sulphur) to form ATP.
2. Both the processes convert nutrients from sugars to form ATP but they differ in their
processes and the amount of energy (ATP) produced.
3. Respiration produces 38 ATP molecules while fermentation produces only 2 ATP molecules..
Determine the intervals of the domain over which the function is cont.pdf
Determine the intervals of the domain over which the function is continuous.
Solution
1)
you can take any value in \"x\", the the answer is the letter (D)
2)
you can\'t take x = 1 because it is indeterminate, so the interval is: (-INF,1) U (1,INF)
So the correct answer is the letter (C)
3)
4)
you can take any value in \"x\", the the answer is the letter (C).
comparison or words in group.F. Nucleotide, NucleusG. Mitosis, C.pdf
comparison or words in group.
F. Nucleotide, Nucleus
G. Mitosis, Cytokinesis
H. Apoptosis, Casas especially
I. Kinases , Cyclins
Solution
Nucleus is the central part of the cell and is round in shape and is most important part which
forms the basis for the activity and growth of the cell. It is a membrane enclosed organelle in
eukaryotic cells and its main function is storing hereditary material (DNA), acts as site of DNA
replication and transcription, ribosome formation.
DNA is a polymer made up of long chain of sub units called nucleotides. Nucleotide is made up
of one sugar (In DNA it is deoxyribose and in RNA it is ribose), one phosphate group and one
nitrogenous group.
Mitosis is a process, where the nuclear division will occur so that the parent cell can divide to
produce two identical daughter cells and it consists of four basic phases: prophase, metaphase,
anaphase, and telophase.
Cytokinesis is the division of the cytoplasm and organelles among the two daughter cells and it is
observed in animal cells, that an actin microfilaments will form a circle between the two
daughter nuclei during the separation process and it will continue to contract finally forming a
furrow ring and it will then squeezes the two daughter cells tighter leading them to separate.
Apoptosis and necrosis are the two ways through which a cell can die. Apoptosis is known as
programmed cell death.
Here a set of proteolytic enzymes called caspasces will cleave specific proteins present in the
nucleus and cytoplasm triggering the cell death.
These caspasces exist in all cells as inactive precursors in normal condition in the form of
procaspases. During apoptosis, these will be activated by other caspasces through cascading
reactions and will begin the process.
Both kinases and clyclins play an important role in cell cycle regulation.
Cdks (Cyclin-Dependent Protein Kinases) through phosphorylation will signal the cell to go to
the next stage of the cell cycle.
But these Cdks are depedent on cyclones, these Cyclins bind to Cdks and activates them.
These play an important role in regulating the cell cycle and also play an important role in
regulating transcription, mRNA processing, and the differentiation of nerve cells.
Nucleus is the central part of the cell and is round in shape and is most important part which
forms the basis for the activity and growth of the cell. It is a membrane enclosed organelle in
eukaryotic cells and its main function is storing hereditary material (DNA), acts as site of DNA
replication and transcription, ribosome formation.
DNA is a polymer made up of long chain of sub units called nucleotides. Nucleotide is made up
of one sugar (In DNA it is deoxyribose and in RNA it is ribose), one phosphate group and one
nitrogenous group.
Mitosis is a process, where the nuclear division will occur so that the parent cell can divide to
produce two identical daughter cells and it consists of four basic phases: prophase, metaphase,
anaphase, and telophase..
Before studying the skin model use your textbook and lab manual to co.pdf
Before studying the skin model use your textbook and lab manual to color and label Figure 4.1.
Epidermal layers (thick skin); high power (xs = cross-sectional view)
Solution
Epidermal identification:
1. Stratum corneum
2. Stratum lucidum.
3. Stratum granulosum.
4. Stratum Spinosum
5. Melanocyte.
Answer the following questions pleaseSolution 5.OPTION A).pdf
Answer the following questions please
Solution
5.
OPTION A)
Listening to music while studying does not affect one\'s learning.
[Null hypotheses almost always says that there is no relationship, no significant evidence, etc.]
6.
OPTION A) a personal probability,
because he is estimating his own ability only, regardless of the performance of others..
Add these three functions to the class binaryTreeType (provided).W.pdf
Add these three functions to the class binaryTreeType (provided).
Write the definition of the function, nodeCount, that returns the number of nodes in the binary
tree.
Write the definition of the function, leavesCount, that takes as a parameter a pointer to the root
node of a binary tree and returns the number of leaves in a binary tree.
Write a function, swapSubtrees, that swaps all of the left and right subtrees of a binary tree. Print
the original tree and the resulting tree using a pre-order traversal.
Write a program to test your new functions. Use the data provided to build your binary search
tree (-999 is the sentinel): (C++)
Data: 65 55 22 44 61 19 90 10 78 52 -999
Already completed code is below:
//Header File Binary Search Tree
#ifndef H_binarySearchTree
#define H_binarySearchTree
#include \"binaryTree.h\"
#include
using namespace std;
template
class bSearchTreeType: public binaryTreeType
{
public:
bool search(const elemType& searchItem) const;
//Function to determine if searchItem is in the binary
//search tree.
//Postcondition: Returns true if searchItem is found in
// the binary search tree; otherwise,
// returns false.
void insert(const elemType& insertItem);
//Function to insert insertItem in the binary search tree.
//Postcondition: If there is no node in the binary search
// tree that has the same info as
// insertItem, a node with the info
// insertItem is created and inserted in the
// binary search tree.
void deleteNode(const elemType& deleteItem);
//Function to delete deleteItem from the binary search tree
//Postcondition: If a node with the same info as deleteItem
// is found, it is deleted from the binary
// search tree.
// If the binary tree is empty or deleteItem
// is not in the binary tree, an appropriate
// message is printed.
private:
void deleteFromTree(nodeType* &p);
//Function to delete the node to which p points is
//deleted from the binary search tree.
//Postcondition: The node to which p points is deleted
// from the binary search tree.
};
template
bool bSearchTreeType::search
(const elemType& searchItem) const
{
nodeType *current;
bool found = false;
if (this->root == NULL)
cout << \"Cannot search an empty tree.\" << endl;
else
{
current = this->root;
while (current != NULL && !found)
{
if (current->info == searchItem)
found = true;
else if (current->info > searchItem)
current = current->lLink;
else
current = current->rLink;
}//end while
}//end else
return found;
}//end search
template
void bSearchTreeType::insert
(const elemType& insertItem)
{
nodeType *current; //pointer to traverse the tree
nodeType *trailCurrent; //pointer behind current
nodeType *newNode; //pointer to create the node
newNode = new nodeType;
newNode->info = insertItem;
newNode->lLink = NULL;
newNode->rLink = NULL;
if (this->root == NULL)
this->root = newNode;
else
{
current = this->root;
while (current != NULL)
{
trailCurrent = current;
if (current->info == insertItem)
{
cout << \"The item to be inserted is already \";
cout << \"in the t.
4. Describe using examples the main characteristics of infrastructure.pdf
4. Describe using examples the main characteristics of infrastructure supply.
Solution
Infrastructure supply includes economic infrastructure and social infrastructure. Economic
infrastructure transportation facilities like rail, road, and airports; water supply, wastewater
treatments; power supply and telecommunication facilities. The main characteristics are the
capital investment, the durability of the project, environmental impacts, type of project whether it
is private, public or public-private partnership etc; usefulness of the project etc..
1. How does the ripened condition of an ovary wall allow you to dete.pdf
1. How does the ripened condition of an ovary wall allow you to determine the category of
fruits?
2. In phylogenetics, what are sister groups? Synapomorphies? Clades? Nodes?
3. How do inflorescences differ?
4. What is the Linnaean system of classification and what are the appropriate endings for
common ranks?
5. Why do gymnosperms lack \"fruits\"?
6. What are synapomorphies and how are they indicated on a phylogeny?
7. Why is the development of seeds considered to be a major advance in the evolution of plants?
Solution
In ripened condition , an ovary develops into a fruit, the ovary wall becomes the pericarp.
============================================
2. In phylogenetics, what are sister groups? Synapomorphies? Clades? Nodes?
Clades : Two clades are sisters if they have an immediate common ancestor.
3. How do inflorescences differ?
inflorescence :it is a group of flowers arranged on a stem that is composed of a main branch .
4What is the Linnaean system of classification and what are the appropriate endings for common
ranks?
In the scientific classification of Linnaeus there are three kingdoms, isolated into classes,
and they, thusly, into
a term for rank-based characterization of living beings, as a rule.
5. Why do gymnosperms lack \"fruits\"?
As the Gymnosperms do not produce flowers and ovaries
synapomorphies :
it is a characteristic present in a genealogical species
Itis shared exclusively.
It is in more or less modified form) by its evolutionary descendants.
synapomorphies. : a character or attribute that is shared by at least two scientific categorizations
and is gotten through advancement from a typical genealogical shape.
7. Why is the development of seeds considered to be a major advance in the evolution of plants?
The development of the seed plants speaks to a noteworthy progress over the more primitive
Bryophytes and greeneries. In plants and greeneries, single- celled, haploid spores are the
essential means for attacking new living spaces. Spores have almost no sustenance stockpiling
and once sprouted can\'t endure extraordinary changes in natural conditions. Additionally, both
greeneries and greeneries require water for preparation while seed plants depend on air streams
and bugs. The seed plants can finish their life cycle completely on dry land, the length of
satisfactory soil dampness from the roots is accessible. Preparation happens inside defensive
tissues of the sporophyte (called integuments) and requires no outside water. Seed- bearing plants
create megaspores (expansive spores) in megasporangia that form into ovules and, at
development, deliver seed. They likewise deliver microspores (little spores) in microsporangia
that form into dust grains (male gametophytes) that convey the sperm to the female structures to
fulfill fertilization. Fertilization must go before preparation..
20. For an independent-samples t test, there were 14 participants in.pdf
20. For an independent-samples t test, there were 14 participants in Group 1 and 17 participants
in Group 2. The total degrees of freedom were: a. 31 b. 29 c. 30 d. 28
Solution
For pooled variance case, df = (n1+nn2-2) = 14+17-2 = 29
Without pooling, conservative approach says df = minimum of (n1,n2)-1 = 13.
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
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Match the graph of the sine function to the equation over [-4 Pi. 4 P.pdf
Match the graph of the sine function to the equation over [-4 Pi. 4 Pi]
Solution
here vertical shift is 1/2
from the graph
Amplitude = 3/2
period = =pi
phase shift is = -pi/4/pi= -1/4
so we can write Y = Asin(BX+C) + D
where A= amplitude
period = 2pi/B
phase shift = -C/B
vertical shift (D)=1/2
so y = 3/2sin(1/2 x -pi/4) +1/2
so it is option A
you can find answer in other way also
from the graph take some points
option A is the correct answer
plug x =-pi/4 then we get y=-1 and in the graph also y=-1.
location in the stem, function, and Chemical Composition of their Wa.pdf
location in the stem, function, and Chemical Composition of their Wall:
-Epidermis
-Periderm
-Xylem
- Phloem
- Parenchyma
- Collenchyma
- Sclerenchyma
Solution
Epidermis-
It is the single layer cell covering which makes the dermal tissue and covers the stem. It is the
outermost layer of the primary plant body.
Function- regulates exchange of gases in plants, prevents the loss of water, and helps in secreting
various metabolic compounds.
Chemical composition- hydrocarbon polymer impregnated with wax and lipids
Periderm-
Periderm is the corky outer layer of the stem. It is the secondary protective dermal tissue which
replaces epidermis during the growth thickness of the stem. Periderm is the multilayer tissue
system.
Function- Protects the stem from pathogen attack an dessication
Chemical composition- polysaccharides, tannins, biopolymers
Xylem-
Xylem is a type of transport tissue located in the vascular bundles.
Function-main function of xylem is to transport water from roots to shoots and leaves along with
some important nutrients.
Chemical composition- hemicelluloses, ligin, cellulose, pectin
Phloem-
Phloem is the living tissue found in the vascular bundles of the plants. It is the innermost layer of
the bark.
Function- the important function is to transport organic compounds formed during the process of
photosynthesis..
If Zn Binomial (n, ) , and n~possion(), find the distribution of Z .pdf
If Z|n Binomial (n, ) , and n~possion(), find the distribution of Z .
Is it more or less dispersed than the distribution of Z|n ?
Solution
since we have a binomial distribution with n and pi
and n has a poisson distribution
Z will have a normal distribution with mean = n by landha
it is more dispersed bvecause n depends of landha.
Given the global financial crisis of 2007–2009, do you anticipate an.pdf
Given the global financial crisis of 2007–2009, do you anticipate any changes to the systems of
fixed exchange rates and forward contracts in the near future? Are the changes you envision
strictly procedural or regulatory, or do you believe that some of these changes will intend to
provide safeguards against ethical lapses or loopholes?
Solution
As a result of the global financial crisis of 2007-2009, it showed that the protection companies
and investors have at the time to hedge against exchange rates instability were severely lacking
resulting in billions of dollars in lost wealth globally. Any financial market carries risks for its
users, but the main function of some markets is redistribution of risk. This is exactly what
forward markets and derivative markets do. They are more risky, but at the same time it is where
risks may be reduced or eliminated. Since the emergence of derivatives market, hedging
operations are increasingly performed in the futures and forward markets (due to minor
transaction costs, more expedient transactions, etc). Investors use forwards with a view to
protecting a certain yield level through the transfer of risk to other trade participants. This risk is
undertaken by speculators who take opposite positions from the hedgers and thus make the
market liquid. Hedgers, however, protect themselves by the transfer of risk to other participants
(speculators) against losses, but also reduce the possibility of profit increase.
Financial crises have many common features. In the background real economy, there is usually
the presence of an asset price “bubble” (or asset price inflation, for purists), a corresponding
credit boom, and large capital inflows into that economy (see, for example, Reinhart and Rogoff,
2008). However, these characteristics are necessary, but not sufficient, for a financial crisis to
develop. The severity of the crisis depends crucially on the underlying financial sector’s
exposure to these conditions and, in fact, the overall market’s uncertainty about the financial
sector’s exposure to them. A key role of financial regulation is to put limits on financial
institutions, so as to limit this exposure. While there are many reasons for the relative calm of the
U.S. financial system during the 50 years after the Great Depression, many analysts continue to
give credit to the financial regulation that was enacted at that time.
Given their inherently high leverage and the ease with which the risk-profile of financial assets
can be altered, banks and financial institutions have incentives to take on excessive risks.
Ordinarily, market mechanisms would be expected to price risks correctly and thereby ensure
that risk-taking in the economy is at efficient levels. However, there are two factors that have
impeded such efficient outcomes. First, with the repeal of most protections from the Banking Act
of the 1933, the only remaining protection against risk-shifting is capital requirements. If the
guarantees are m.
Generativity versus stagnation Early adult transitio.pdf
Generativity versus stagnation
Early adult transition
Intimacy versus isolation
Self-awareness transition
Generativity involves a concern for and interest in establishing and guiding the next
generation.
Generativity involves a commitment to improve the life conditions of future generations.
Generativity involves a commitment towards protecting and enhancing the conditions of
one’s society.
All of the above are true.
Socializing versus sexualizing in human relationships
Emotional flexibility versus emotional impoverishment
Culminating life structure versus early adult transition
Valuing wisdom versus valuing physical powers
Sexualizing in human relationships
Valuing physical powers
Emotional impoverishment
Mental flexibility
Early adult transition
Entry life structure for early adulthood
Culminating life structure for early adulthood
Entry life structure for middle adulthoodErikson theorized that after young people develop a
sense of identity, they next face the psychosocial crisis of: (Points : 2)
Solution
Soln:
1) Intimacy versus isolation
2) All of the above are true.
3) Culminating life structure versus early adult transition
4) Mental flexibility
5) Early adult transition
6) Entry life structure for middle adulthood.
Genetics Question 2 In the insect sex determination system, th.pdf
Genetics
Question 2 In the insect sex determination system, the are considered the homogametic sex
because they have 2X chromosomes females males
Solution
In insects, mostly the sex determination is of XX-XY type. In this, females are homogametic i.e.
they contain two copies of the same kind of sex chromosomes. Females contain two X
chromosomes. Males are heterogametic i.e. they contain two different kinds of sex
chromosomes. Hence, males contain one X chromosome and one Y chromosome.
Therefore, the correct answer is females..
Explain why malaria infection may lead to anemia. How would you clas.pdf
Explain why malaria infection may lead to anemia. How would you classify the anemia that
results?(scientific name for the type of anemia that results)
Solution
Plasmodium falciperum and Plasmodium vivax causes malaria in human and also responsible for
anemia .It is due to increase removal of erythrocyte as well as decreased production of
erythrocytes in the the bone marrow.Anemia is due to the continous removal/ destruction of
erythrocytes in the bone marrow. Parasitie infect RBC and its toxin causes harmful effect to the
erythrocyte. It causes severe anemia in human..
Figure 13.2 of a single pair of homologous chromosomes as they might.pdf
Figure 13.2 of a single pair of homologous chromosomes as they might appear during various
stages of either mitosis or meiosis 53) Which diagram(s) represent anaphase II of meiosis? A) II
only B) III only C) IV only D) V only E) either II or V Answer: D What\'s wrong with E?? If III
V VI /1 /N
Solution
It Showing various stages of Meiosis because the stages showing 2 anaphases, crossing over and
chiasma formation
Yaa the answer is D. Anaphase 2 occur in haploid cells, so it contain single chromosome but 2
chromatids. The two chromatids seperate in the Anaphase II. So D is correct.
Every individual produced by sexual reproduction is unique. This is .pdf
Every individual produced by sexual reproduction is unique. This is because of… (circle the
TWO correct answers)
a) Independent assortment of chromosomes during meiosis.
b) Independent assortment of chromosomes during mitosis.
c) Recombination occurring in prophase.
d) Separation of chromatids occurring in anaphase 2.
Solution
a. Independent assortment of chromosomes during Meiosis
c. Recombination occuring in prophase.
After sexual reproduction the gametes that unite to form zygote are haploid. Since the haploid
gametes are formed by the reductional divisions of diploid cell, So, The chromosomes that
haploid gamete will carry, will be responsible for the characters inherited to the progeny
Also the recombination that takes place in the prophase, determines the characters or traits
inherited after the crossing over of chromosomes.
Thus these two factors will be most responsible in determining the individuality of the progeny..
Explain to them the probability of future offspring being normal or h.pdf
Explain to them the probability of future offspring being normal or having the disease (assume
that this is an autosomal disease). Be sure to tell them whether the disease is brought about by a
dominant or recessive allele and what fraction of their children (if any) could be carriers for the
disease. Be sure to incorporate the following into your explanation: DNA and chromosomes,
genes, dominant and recessive alleles, paired alleles on homologous chromosomes, and
segregation of alleles during gamete production. You can use words and drawings/diagrams.
Sally and Steve are concerned about cystic fibrosis. Neither of them has it, nor do their parents or
grandparents, yet each one has a sibling with cystic fibrosis (Sally\'s sister and Steve\'s brother).
Sally and Steve want to have a second child (the first one, age 4, appears unaffected). Jess and
Jessie want to get married and have children. Unfortunately, they have the same great-
grandmother (not the same great-grandfather). The great-grandmother suffered from polycystic
kidney disease. No other family members suffered from the disease, but they were told
(correctly) that their great-grandmother had passed on the possibility of the disease.
Solution
Ques-1:
Aetiology and Pathophysiology: It is autosomal recessive disease (mode of inheritance), which is
familial. Gene for cystic fibrosis is located on the chromosome 7 & this recessive allele
propogated to the offspring via parental gamete formation by reduction division & problem will
be at segregation of chromosomes. A protein called, CF transmembrane -this gene produces
regulator (CFTR). In case of Pleiotrophy one affects 2 or more apparently not linked phenotypic
traits there by more gene could be affected (cystic fibrosis).
For instance, Sally and Steve are concerned about cystic fibrosis. Neither of them has it, nor do
their parents or grandparents, yet each one has a sibling with cystic fibrosis (Sally\'s sister and
Steve\'s brother) i.e. both Sally & Steve are carriers of the disease. Therefore, 2/3 (67%), of
probability with unaffected child. Sally and Steve want to have a second child but the first child
age-4 has no cystic fibrosis
Cystic fibrosis is an autosomal recessive condition. In the given example, the unaffected parents
will have an affected children Assume that the gene coding for the normal allele is C, and the
gene coding for the cystic fibrosis is c and C is dominant over c. Now, cross between two
heterozygotes will have the offspring with the following genotypes.
Cc* Cc = CC, Cc, Cc, cc (means, 3/4th are unaffected (out of these 2/4 are carriers) and the ¼
are affected -having disease).
C
c
C
CC
Cc
c
Cc
cc
C
c
C
CC
Cc
c
Cc
cc.
explain conceptual questionsvariabilitystandard devationz-sc.pdf
explain conceptual questions
variability?
standard devation?
z-scores (include a detailed drawing of a sample z-distribution)?
what is factorial design and why do we use them in psychology?
hypothesis testing( explain the null and alternative hypotheses- which one do we test against in
order to determin hte siginficance, alpha levels, and how probability is used to determin
significance)?
Solution.
Explain the difference between fermentation and respiration in biolo.pdf
Explain the difference between fermentation and respiration in biology?
Solution
Fermentation and Respiration are the two widely used processes in biology.
Respiration : It is a biochemical process in which food molecules are broken down by cells of an
organism with the release of carbon dioxide , water and ATP ( energy currency of cells).
Food + oxygen gives carbondioxide + water + ATP.
Types : 1. Aerobic respiration : Occurs in presence of oxygen, performed by eukaryotic cells,
yields 38 ATP molecules and Occurs in both cytosol and mitochondria.
2 : Anaerobic Respiration : Occurs in absence of oxygen, performed by prokaryotic cells, yields
only 2 ATP molecules and Occurs in cytosol only.
Fermentation : Term ist used by Louis Pasteur in 19th century.
It is a metabolic process in which sugar is converted to acids or alcohol.
It occurs in yeast and bacteria and sometimes in muscle cells ( lactic acid fermentation )
Fermentation Occurs when electron transport chain becomes defunct due to lack of final electron
acceptor I,e oxygen.
Difference between Respiration and Fermentation:
1. Respiration uses oxygen as electron acceptor to form ATP while fermentation uses inorganic
donars ( Sulphur) to form ATP.
2. Both the processes convert nutrients from sugars to form ATP but they differ in their
processes and the amount of energy (ATP) produced.
3. Respiration produces 38 ATP molecules while fermentation produces only 2 ATP molecules..
Determine the intervals of the domain over which the function is cont.pdf
Determine the intervals of the domain over which the function is continuous.
Solution
1)
you can take any value in \"x\", the the answer is the letter (D)
2)
you can\'t take x = 1 because it is indeterminate, so the interval is: (-INF,1) U (1,INF)
So the correct answer is the letter (C)
3)
4)
you can take any value in \"x\", the the answer is the letter (C).
comparison or words in group.F. Nucleotide, NucleusG. Mitosis, C.pdf
comparison or words in group.
F. Nucleotide, Nucleus
G. Mitosis, Cytokinesis
H. Apoptosis, Casas especially
I. Kinases , Cyclins
Solution
Nucleus is the central part of the cell and is round in shape and is most important part which
forms the basis for the activity and growth of the cell. It is a membrane enclosed organelle in
eukaryotic cells and its main function is storing hereditary material (DNA), acts as site of DNA
replication and transcription, ribosome formation.
DNA is a polymer made up of long chain of sub units called nucleotides. Nucleotide is made up
of one sugar (In DNA it is deoxyribose and in RNA it is ribose), one phosphate group and one
nitrogenous group.
Mitosis is a process, where the nuclear division will occur so that the parent cell can divide to
produce two identical daughter cells and it consists of four basic phases: prophase, metaphase,
anaphase, and telophase.
Cytokinesis is the division of the cytoplasm and organelles among the two daughter cells and it is
observed in animal cells, that an actin microfilaments will form a circle between the two
daughter nuclei during the separation process and it will continue to contract finally forming a
furrow ring and it will then squeezes the two daughter cells tighter leading them to separate.
Apoptosis and necrosis are the two ways through which a cell can die. Apoptosis is known as
programmed cell death.
Here a set of proteolytic enzymes called caspasces will cleave specific proteins present in the
nucleus and cytoplasm triggering the cell death.
These caspasces exist in all cells as inactive precursors in normal condition in the form of
procaspases. During apoptosis, these will be activated by other caspasces through cascading
reactions and will begin the process.
Both kinases and clyclins play an important role in cell cycle regulation.
Cdks (Cyclin-Dependent Protein Kinases) through phosphorylation will signal the cell to go to
the next stage of the cell cycle.
But these Cdks are depedent on cyclones, these Cyclins bind to Cdks and activates them.
These play an important role in regulating the cell cycle and also play an important role in
regulating transcription, mRNA processing, and the differentiation of nerve cells.
Nucleus is the central part of the cell and is round in shape and is most important part which
forms the basis for the activity and growth of the cell. It is a membrane enclosed organelle in
eukaryotic cells and its main function is storing hereditary material (DNA), acts as site of DNA
replication and transcription, ribosome formation.
DNA is a polymer made up of long chain of sub units called nucleotides. Nucleotide is made up
of one sugar (In DNA it is deoxyribose and in RNA it is ribose), one phosphate group and one
nitrogenous group.
Mitosis is a process, where the nuclear division will occur so that the parent cell can divide to
produce two identical daughter cells and it consists of four basic phases: prophase, metaphase,
anaphase, and telophase..
Before studying the skin model use your textbook and lab manual to co.pdf
Before studying the skin model use your textbook and lab manual to color and label Figure 4.1.
Epidermal layers (thick skin); high power (xs = cross-sectional view)
Solution
Epidermal identification:
1. Stratum corneum
2. Stratum lucidum.
3. Stratum granulosum.
4. Stratum Spinosum
5. Melanocyte.
Answer the following questions pleaseSolution 5.OPTION A).pdf
Answer the following questions please
Solution
5.
OPTION A)
Listening to music while studying does not affect one\'s learning.
[Null hypotheses almost always says that there is no relationship, no significant evidence, etc.]
6.
OPTION A) a personal probability,
because he is estimating his own ability only, regardless of the performance of others..
Add these three functions to the class binaryTreeType (provided).W.pdf
Add these three functions to the class binaryTreeType (provided).
Write the definition of the function, nodeCount, that returns the number of nodes in the binary
tree.
Write the definition of the function, leavesCount, that takes as a parameter a pointer to the root
node of a binary tree and returns the number of leaves in a binary tree.
Write a function, swapSubtrees, that swaps all of the left and right subtrees of a binary tree. Print
the original tree and the resulting tree using a pre-order traversal.
Write a program to test your new functions. Use the data provided to build your binary search
tree (-999 is the sentinel): (C++)
Data: 65 55 22 44 61 19 90 10 78 52 -999
Already completed code is below:
//Header File Binary Search Tree
#ifndef H_binarySearchTree
#define H_binarySearchTree
#include \"binaryTree.h\"
#include
using namespace std;
template
class bSearchTreeType: public binaryTreeType
{
public:
bool search(const elemType& searchItem) const;
//Function to determine if searchItem is in the binary
//search tree.
//Postcondition: Returns true if searchItem is found in
// the binary search tree; otherwise,
// returns false.
void insert(const elemType& insertItem);
//Function to insert insertItem in the binary search tree.
//Postcondition: If there is no node in the binary search
// tree that has the same info as
// insertItem, a node with the info
// insertItem is created and inserted in the
// binary search tree.
void deleteNode(const elemType& deleteItem);
//Function to delete deleteItem from the binary search tree
//Postcondition: If a node with the same info as deleteItem
// is found, it is deleted from the binary
// search tree.
// If the binary tree is empty or deleteItem
// is not in the binary tree, an appropriate
// message is printed.
private:
void deleteFromTree(nodeType* &p);
//Function to delete the node to which p points is
//deleted from the binary search tree.
//Postcondition: The node to which p points is deleted
// from the binary search tree.
};
template
bool bSearchTreeType::search
(const elemType& searchItem) const
{
nodeType *current;
bool found = false;
if (this->root == NULL)
cout << \"Cannot search an empty tree.\" << endl;
else
{
current = this->root;
while (current != NULL && !found)
{
if (current->info == searchItem)
found = true;
else if (current->info > searchItem)
current = current->lLink;
else
current = current->rLink;
}//end while
}//end else
return found;
}//end search
template
void bSearchTreeType::insert
(const elemType& insertItem)
{
nodeType *current; //pointer to traverse the tree
nodeType *trailCurrent; //pointer behind current
nodeType *newNode; //pointer to create the node
newNode = new nodeType;
newNode->info = insertItem;
newNode->lLink = NULL;
newNode->rLink = NULL;
if (this->root == NULL)
this->root = newNode;
else
{
current = this->root;
while (current != NULL)
{
trailCurrent = current;
if (current->info == insertItem)
{
cout << \"The item to be inserted is already \";
cout << \"in the t.
4. Describe using examples the main characteristics of infrastructure.pdf
4. Describe using examples the main characteristics of infrastructure supply.
Solution
Infrastructure supply includes economic infrastructure and social infrastructure. Economic
infrastructure transportation facilities like rail, road, and airports; water supply, wastewater
treatments; power supply and telecommunication facilities. The main characteristics are the
capital investment, the durability of the project, environmental impacts, type of project whether it
is private, public or public-private partnership etc; usefulness of the project etc..
1. How does the ripened condition of an ovary wall allow you to dete.pdf
1. How does the ripened condition of an ovary wall allow you to determine the category of
fruits?
2. In phylogenetics, what are sister groups? Synapomorphies? Clades? Nodes?
3. How do inflorescences differ?
4. What is the Linnaean system of classification and what are the appropriate endings for
common ranks?
5. Why do gymnosperms lack \"fruits\"?
6. What are synapomorphies and how are they indicated on a phylogeny?
7. Why is the development of seeds considered to be a major advance in the evolution of plants?
Solution
In ripened condition , an ovary develops into a fruit, the ovary wall becomes the pericarp.
============================================
2. In phylogenetics, what are sister groups? Synapomorphies? Clades? Nodes?
Clades : Two clades are sisters if they have an immediate common ancestor.
3. How do inflorescences differ?
inflorescence :it is a group of flowers arranged on a stem that is composed of a main branch .
4What is the Linnaean system of classification and what are the appropriate endings for common
ranks?
In the scientific classification of Linnaeus there are three kingdoms, isolated into classes,
and they, thusly, into
a term for rank-based characterization of living beings, as a rule.
5. Why do gymnosperms lack \"fruits\"?
As the Gymnosperms do not produce flowers and ovaries
synapomorphies :
it is a characteristic present in a genealogical species
Itis shared exclusively.
It is in more or less modified form) by its evolutionary descendants.
synapomorphies. : a character or attribute that is shared by at least two scientific categorizations
and is gotten through advancement from a typical genealogical shape.
7. Why is the development of seeds considered to be a major advance in the evolution of plants?
The development of the seed plants speaks to a noteworthy progress over the more primitive
Bryophytes and greeneries. In plants and greeneries, single- celled, haploid spores are the
essential means for attacking new living spaces. Spores have almost no sustenance stockpiling
and once sprouted can\'t endure extraordinary changes in natural conditions. Additionally, both
greeneries and greeneries require water for preparation while seed plants depend on air streams
and bugs. The seed plants can finish their life cycle completely on dry land, the length of
satisfactory soil dampness from the roots is accessible. Preparation happens inside defensive
tissues of the sporophyte (called integuments) and requires no outside water. Seed- bearing plants
create megaspores (expansive spores) in megasporangia that form into ovules and, at
development, deliver seed. They likewise deliver microspores (little spores) in microsporangia
that form into dust grains (male gametophytes) that convey the sperm to the female structures to
fulfill fertilization. Fertilization must go before preparation..
20. For an independent-samples t test, there were 14 participants in.pdf
20. For an independent-samples t test, there were 14 participants in Group 1 and 17 participants
in Group 2. The total degrees of freedom were: a. 31 b. 29 c. 30 d. 28
Solution
For pooled variance case, df = (n1+nn2-2) = 14+17-2 = 29
Without pooling, conservative approach says df = minimum of (n1,n2)-1 = 13.
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KKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKKK
Operation Blue Star - Saka Neela Tara
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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