If you built your observatory at 45N, could you see stars at declina.pdf
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If you built your observatory at 45N, could you see stars at declination -10 Solution No, because the Earth rotates in a circle. These star trails exist because of the Earth\'s rotation, so if you were to look towards the North Star, it wouldn\'t move noticeably over time, as it is at the pole and the stars seem to orbit around that axis. The stars do move though, orbiting around the center of the galaxy, but it is a very slow process and the stars close to us move at around the same speed, so it\'s not noticeable..
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During January 2005, an Italian investor investing in the Italian st.pdf
During January 2005, an Italian investor investing in the Italian stock market earned a return of
1.47%. During the same month, an American investor investing in the Italian stock market
earned a return of –2.358%. The U.S. dollar-Euro exchange rate at the end of January 2005 was
1.3035 $/€. What was the U.S. dollar-Euro exchange rate ($/€) at the end of December 2004?
Solution
F = S * ((1 + if) / (1 + id)) 1.3035 = S x (1+1.47%) / (1 - 2.358%) S = U.S. dollar-Euro
exchange rate ($/€) = $1.25.
Distinguish between protostomes and deuterostomes. Identify the Phyla.pdf
Protostomes develop with the mouth forming from the blastopore, and include phyla like flatworms, nematodes, mollusks, annelids, and arthropods. Deuterostomes have the mouth form from a second opening on the dorsal surface while the blastopore becomes the anus, and include the phyla Chordata, Echinodermata, and Hemichordata. Deuterostomes also exhibit indeterminate development where cells remain capable of forming a complete organism even when separated, unlike determinate development where each cell's fate is predetermined.
Determine the y-coordinate of the centroid of the shaded area. S.pdf
Determine the y-coordinate of the centroid of the shaded area.
Solution
Ans-
General topology grew out of a number of areas, most importantly the following:
General topology assumed its present form around 1940. It captures, one might say, almost
everything in the intuition of continuity, in a technically adequate form that can be applied in any
area of mathematics.
A topology on a set
Main article: Topological space
Let X be a set and let be a family of subsets of X. Then is called a topology on X if:[1][2]
If is a topology on X, then the pair (X, ) is called a topological space. The notation X may be
used to denote a set X endowed with the particular topology .
The members of are called open sets in X. A subset of X is said to be closed if its complement is
in (i.e., its complement is open). A subset of X may be open, closed, both (clopen set), or
neither. The empty set and X itself are always both closed and open.
Basis for a topology
Main article: Basis (topology)
A base (or basis) B for a topological space X with topology T is a collection of open sets in T
such that every open set in T can be written as a union of elements of B.[3][4] We say that the
base generates the topology T. Bases are useful because many properties of topologies can be
reduced to statements about a base that generates that topology—and because many topologies
are most easily defined in terms of a base that generates them.
Subspace and quotient
Every subset of a topological space can be given the subspace topology in which the open sets
are the intersections of the open sets of the larger space with the subset. For any indexed family
of topological spaces, the product can be given the product topology, which is generated by the
inverse images of open sets of the factors under the projection mappings. For example, in finite
products, a basis for the product topology consists of all products of open sets. For infinite
products, there is the additional requirement that in a basic open set, all but finitely many of its
projections are the entire space.
A quotient space is defined as follows: if X is a topological space and Y is a set, and if f : XY is
a surjective function, then the quotient topology on Y is the collection of subsets of Y that have
open inverse images under f. In other words, the quotient topology is the finest topology on Y for
which f is continuous. A common example of a quotient topology is when an equivalence
relation is defined on the topological space X. The map f is then the natural projection onto the
set of equivalence classes..
Describe the mechanism by which TLR4, CD14, and MD2 recognize Lipopo.pdf
Describe the mechanism by which TLR4, CD14, and MD2 recognize Lipopolysaccharides (LPS)
and how this recognition changes gene expression in macrophages. What would happen if the
TIR domain was mutated?
Solution
LPS is a main component of the outer membrane of Gram-negative bacteria. Minute amounts of
lipopolysaccharides released from infecting pathogens can initiate potent innate immune
responses. But when the Lipopolysaccharides response is not properly controlled it can lead to
fatal septic shock syndrome. The common structural pattern of Lipopolysaccharides in various
bacterial species is recognized by LPS binding protein (LBP), CD14 and the Toll-like receptor4
(TLR4)–MD-2 complex. Lipopolysaccharide is initially extracted from bacterial membranes and
vesicles released from them by LPS binding protein in serum. LPS binding protein then transfers
LPS to CD14 that can be found either in soluble form or linked to the cell surface by a
glycosylphosphatidylinositol anchor. CD14 splits Lipopolysaccharides aggregates into
monomeric molecules and presents them to the TLR4–MD-2 complex. Aggregation of the
TLR4–MD-2 complex after binding lipopolysaccharide leads to activation of multiple signaling
components like NF-B and IRF3, and the subsequent production of pro-inflammatory cytokines.
CD14 exists in the form of a homodimer. Unlike other LRR family members, it does not contain
an LRRCT module. Instead, the C-terminus of the LRR modules of one CD14 molecule interacts
with the C-terminus of another, resulting in the formation of a dimer. The lipopolysaccharide
interaction pocket of CD14 is located at the boundary of the LRRNT and the first LRR module.
In addition to LPS, CD14 can bind to various other microbial products, such as peptidoglycan,
lipoteichoic acid, lipoarabinomannan, and lipoproteins. Hence, it has broad ligand specificity. It
functions as a pattern recognition receptor by recognizing structural motifs in various microbial
products. The extracellular domain of TLR4 also belongs to the LRR family and is responsible
for ligand binding. The structure of TLR4 deviates substantially from the consensus LRR
structures. LRR family proteins can be classified into seven subfamilies and each subfamily is
defined by its conformation. Most of the LRR proteins have uniform radii and beta sheet angles.
Whereas, TLRs 1, 2, 4 and 6 are divided by a structural transition into N-terminal, central and C-
terminal. The structural discontinuities seem to be caused by irregular LRR sequences in the
center. The domain boundaries in TLRs 1, 2 and 6 fulfill important roles in ligand binding. The
N-terminal and central domains of TLR4 clearly provide charge complementarity for binding of
its surface to its co-receptor MD-2 thus, forming a stable 1:1 heterodimer. In the initial phase of
infection, the innate immune system generates a rapid inflammatory response which blocks the
growth of the infectious agent. Such response is followed, in vertebrates, by an adapt.
Be specific..LIST and GIVE AN EXAMPLE of each of the four types of.pdf
Be specific..
LIST and GIVE AN EXAMPLE of each of the four types of market failure...
(i.e. the reasons that we need government intervention)
Solution
The main reasons because of which we need government intervention are as follows:-
1.) Poverty alleviation:- Poverty is one of major problem in underdeveloped and developing
nations. Government runs various poverty alleviation programmes. For example, Prime
Minister\'s Rozgar Yojna (PMRY) in India. Through these programmes, government accelerates
the economic growth. It helps in population control through awareness regarding various family
planning programmes. It tries to reduces income inequalities in the nation.
2) Human capital formation:- It implies development of abilities and skills among the population
of country. It is the process of acquiring and increasing the number of people, who have skills,
experience and education, which are critical for social, economical and political development
and growth of the nation. The state and union governments have been earmarking substantial
financial outlays for development of education and health sectors in the country.
3) Rural development:- Government plays a very vital role in developing rural life through
providing drinking water facility, proper sanitation and severage facility, Basic amenities like
Housing etc. Government aims at improving social and economic conditions of people living in
villages through various awareness programmes and campaigns.
4) Employment Generation:- Union and state governments played vital role in generating
employment or creating opportunities for employment generation. Government provides \'Direct
Employment\' by employing people in various departments for administrative purposes and in
different industries, hotels, transport companies etc. run by the government.
Therefore these are the Four areas ( Poverty alleviation, Human capital formation, Rural
development and Employment generation etc.) which need government intervention to the large
extent..
What is the correct sequence of events for the life cycle of the pork.pdf
What is the correct sequence of events for the life cycle of the pork tapeworm, T. sodium?
Arrange the following statements in chronological order
Solution
Answering the sequence according to ascending order:
1. Eggs are produced by adult tapeworms living in definitive host
2. Eggs are released in the environment in the feces
3. Pigs eat grass contaminated, or fertilized with human feces
4. Inside the intermediate hosst, the eggs mature into larvae and encyst in the host muscles
5. The definitive host ingests the larvae and becomes infected.
Which heterologous protein expression systems allow for the post-tra.pdf
Which heterologous protein expression systems allow for the post-translational modifications of
the recombinant proteins?
Solution
Yeast systems are generally regarded as safe heterologous recombinant (suitable for DNA
recombination) system for the expression of proteins via heterologous expression
(Saccharomyces cerevisiae). This yeast is unicellular eukaryotic system & its secretory pathway
has associated with N-terminal signal sequences such as Pho5, Suc2 for post translational and
co-translational modifications of secretory recombinant proteins (obtained through recombinant
DNA expression) in endoplasmic reticulum. The post-translational modification can be done
through \"signal peptidases\" & N-linked and O-linked (to serine/threonine) glycosylation..
What is the structure of electric circuits Can a branch of a circui.pdf
What is the structure of electric circuits? Can a branch of a circuit contain three elements? Can
there be a branch which doesn’t contain any circuit elements?
Solution
First electric circuit means, the circuit having elements like resistors, capacitors and inductors
along with sources.
Any electric circuit comprises of node, branches, and meshes.
The joining point of two elements is called node.
The element between two nodes is called branch. In branch only one series element is possible to
connect but in parallel it can have N number of elements.
No branch with out elements in electric circuits..
Recommended
During January 2005, an Italian investor investing in the Italian st.pdf
During January 2005, an Italian investor investing in the Italian stock market earned a return of
1.47%. During the same month, an American investor investing in the Italian stock market
earned a return of –2.358%. The U.S. dollar-Euro exchange rate at the end of January 2005 was
1.3035 $/€. What was the U.S. dollar-Euro exchange rate ($/€) at the end of December 2004?
Solution
F = S * ((1 + if) / (1 + id)) 1.3035 = S x (1+1.47%) / (1 - 2.358%) S = U.S. dollar-Euro
exchange rate ($/€) = $1.25.
Distinguish between protostomes and deuterostomes. Identify the Phyla.pdf
Protostomes develop with the mouth forming from the blastopore, and include phyla like flatworms, nematodes, mollusks, annelids, and arthropods. Deuterostomes have the mouth form from a second opening on the dorsal surface while the blastopore becomes the anus, and include the phyla Chordata, Echinodermata, and Hemichordata. Deuterostomes also exhibit indeterminate development where cells remain capable of forming a complete organism even when separated, unlike determinate development where each cell's fate is predetermined.
Determine the y-coordinate of the centroid of the shaded area. S.pdf
Determine the y-coordinate of the centroid of the shaded area.
Solution
Ans-
General topology grew out of a number of areas, most importantly the following:
General topology assumed its present form around 1940. It captures, one might say, almost
everything in the intuition of continuity, in a technically adequate form that can be applied in any
area of mathematics.
A topology on a set
Main article: Topological space
Let X be a set and let be a family of subsets of X. Then is called a topology on X if:[1][2]
If is a topology on X, then the pair (X, ) is called a topological space. The notation X may be
used to denote a set X endowed with the particular topology .
The members of are called open sets in X. A subset of X is said to be closed if its complement is
in (i.e., its complement is open). A subset of X may be open, closed, both (clopen set), or
neither. The empty set and X itself are always both closed and open.
Basis for a topology
Main article: Basis (topology)
A base (or basis) B for a topological space X with topology T is a collection of open sets in T
such that every open set in T can be written as a union of elements of B.[3][4] We say that the
base generates the topology T. Bases are useful because many properties of topologies can be
reduced to statements about a base that generates that topology—and because many topologies
are most easily defined in terms of a base that generates them.
Subspace and quotient
Every subset of a topological space can be given the subspace topology in which the open sets
are the intersections of the open sets of the larger space with the subset. For any indexed family
of topological spaces, the product can be given the product topology, which is generated by the
inverse images of open sets of the factors under the projection mappings. For example, in finite
products, a basis for the product topology consists of all products of open sets. For infinite
products, there is the additional requirement that in a basic open set, all but finitely many of its
projections are the entire space.
A quotient space is defined as follows: if X is a topological space and Y is a set, and if f : XY is
a surjective function, then the quotient topology on Y is the collection of subsets of Y that have
open inverse images under f. In other words, the quotient topology is the finest topology on Y for
which f is continuous. A common example of a quotient topology is when an equivalence
relation is defined on the topological space X. The map f is then the natural projection onto the
set of equivalence classes..
Describe the mechanism by which TLR4, CD14, and MD2 recognize Lipopo.pdf
Describe the mechanism by which TLR4, CD14, and MD2 recognize Lipopolysaccharides (LPS)
and how this recognition changes gene expression in macrophages. What would happen if the
TIR domain was mutated?
Solution
LPS is a main component of the outer membrane of Gram-negative bacteria. Minute amounts of
lipopolysaccharides released from infecting pathogens can initiate potent innate immune
responses. But when the Lipopolysaccharides response is not properly controlled it can lead to
fatal septic shock syndrome. The common structural pattern of Lipopolysaccharides in various
bacterial species is recognized by LPS binding protein (LBP), CD14 and the Toll-like receptor4
(TLR4)–MD-2 complex. Lipopolysaccharide is initially extracted from bacterial membranes and
vesicles released from them by LPS binding protein in serum. LPS binding protein then transfers
LPS to CD14 that can be found either in soluble form or linked to the cell surface by a
glycosylphosphatidylinositol anchor. CD14 splits Lipopolysaccharides aggregates into
monomeric molecules and presents them to the TLR4–MD-2 complex. Aggregation of the
TLR4–MD-2 complex after binding lipopolysaccharide leads to activation of multiple signaling
components like NF-B and IRF3, and the subsequent production of pro-inflammatory cytokines.
CD14 exists in the form of a homodimer. Unlike other LRR family members, it does not contain
an LRRCT module. Instead, the C-terminus of the LRR modules of one CD14 molecule interacts
with the C-terminus of another, resulting in the formation of a dimer. The lipopolysaccharide
interaction pocket of CD14 is located at the boundary of the LRRNT and the first LRR module.
In addition to LPS, CD14 can bind to various other microbial products, such as peptidoglycan,
lipoteichoic acid, lipoarabinomannan, and lipoproteins. Hence, it has broad ligand specificity. It
functions as a pattern recognition receptor by recognizing structural motifs in various microbial
products. The extracellular domain of TLR4 also belongs to the LRR family and is responsible
for ligand binding. The structure of TLR4 deviates substantially from the consensus LRR
structures. LRR family proteins can be classified into seven subfamilies and each subfamily is
defined by its conformation. Most of the LRR proteins have uniform radii and beta sheet angles.
Whereas, TLRs 1, 2, 4 and 6 are divided by a structural transition into N-terminal, central and C-
terminal. The structural discontinuities seem to be caused by irregular LRR sequences in the
center. The domain boundaries in TLRs 1, 2 and 6 fulfill important roles in ligand binding. The
N-terminal and central domains of TLR4 clearly provide charge complementarity for binding of
its surface to its co-receptor MD-2 thus, forming a stable 1:1 heterodimer. In the initial phase of
infection, the innate immune system generates a rapid inflammatory response which blocks the
growth of the infectious agent. Such response is followed, in vertebrates, by an adapt.
Be specific..LIST and GIVE AN EXAMPLE of each of the four types of.pdf
Be specific..
LIST and GIVE AN EXAMPLE of each of the four types of market failure...
(i.e. the reasons that we need government intervention)
Solution
The main reasons because of which we need government intervention are as follows:-
1.) Poverty alleviation:- Poverty is one of major problem in underdeveloped and developing
nations. Government runs various poverty alleviation programmes. For example, Prime
Minister\'s Rozgar Yojna (PMRY) in India. Through these programmes, government accelerates
the economic growth. It helps in population control through awareness regarding various family
planning programmes. It tries to reduces income inequalities in the nation.
2) Human capital formation:- It implies development of abilities and skills among the population
of country. It is the process of acquiring and increasing the number of people, who have skills,
experience and education, which are critical for social, economical and political development
and growth of the nation. The state and union governments have been earmarking substantial
financial outlays for development of education and health sectors in the country.
3) Rural development:- Government plays a very vital role in developing rural life through
providing drinking water facility, proper sanitation and severage facility, Basic amenities like
Housing etc. Government aims at improving social and economic conditions of people living in
villages through various awareness programmes and campaigns.
4) Employment Generation:- Union and state governments played vital role in generating
employment or creating opportunities for employment generation. Government provides \'Direct
Employment\' by employing people in various departments for administrative purposes and in
different industries, hotels, transport companies etc. run by the government.
Therefore these are the Four areas ( Poverty alleviation, Human capital formation, Rural
development and Employment generation etc.) which need government intervention to the large
extent..
What is the correct sequence of events for the life cycle of the pork.pdf
What is the correct sequence of events for the life cycle of the pork tapeworm, T. sodium?
Arrange the following statements in chronological order
Solution
Answering the sequence according to ascending order:
1. Eggs are produced by adult tapeworms living in definitive host
2. Eggs are released in the environment in the feces
3. Pigs eat grass contaminated, or fertilized with human feces
4. Inside the intermediate hosst, the eggs mature into larvae and encyst in the host muscles
5. The definitive host ingests the larvae and becomes infected.
Which heterologous protein expression systems allow for the post-tra.pdf
Which heterologous protein expression systems allow for the post-translational modifications of
the recombinant proteins?
Solution
Yeast systems are generally regarded as safe heterologous recombinant (suitable for DNA
recombination) system for the expression of proteins via heterologous expression
(Saccharomyces cerevisiae). This yeast is unicellular eukaryotic system & its secretory pathway
has associated with N-terminal signal sequences such as Pho5, Suc2 for post translational and
co-translational modifications of secretory recombinant proteins (obtained through recombinant
DNA expression) in endoplasmic reticulum. The post-translational modification can be done
through \"signal peptidases\" & N-linked and O-linked (to serine/threonine) glycosylation..
What is the structure of electric circuits Can a branch of a circui.pdf
What is the structure of electric circuits? Can a branch of a circuit contain three elements? Can
there be a branch which doesn’t contain any circuit elements?
Solution
First electric circuit means, the circuit having elements like resistors, capacitors and inductors
along with sources.
Any electric circuit comprises of node, branches, and meshes.
The joining point of two elements is called node.
The element between two nodes is called branch. In branch only one series element is possible to
connect but in parallel it can have N number of elements.
No branch with out elements in electric circuits..
Turner syndrome is caused by monosomy of the sex chromosomes XO. .pdf
Turner syndrome is caused by monosomy of the sex chromosomes XO. Why is there no
corresponding condition for monosomy of the sex chromosome YO
Turner syndrome is caused by monosomy of the sex chromosomes XO. Why is there no
corresponding condition for monosomy of the sex chromosome YO
Solution
Nondisjunction leading to YO cells are non viable.At least one X chromosome is needed for
viability.
YO condition is lethal hence there are no corresponding conditions for monosomy of the sex
chromosome YO..
why are we seeing more evidence of widespread etchical dilemmas with.pdf
why are we seeing more evidence of widespread etchical dilemmas within organization?
Solution
Social Responsibility and caring and trust worthiness has been started to decrease, the reason is
thinking about more profit with less effort.
Lack of internal cooperation and Misunderstanding between the workers, it was raised..
What do evebitda ratio and price to cash flow ratio show us .pdf
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
Solution
EV/EBITDA ratio (enterprise value divided by earnings before interest, tax, depreciation and
amortization) is an important valuation multiple and the reason why it is so popular is that the
ratio makes adjustments so that an investor can identify the real cash flow issue associated with a
company. The ratio shows the real value of a business and hence can be used to easily compare
companies that operate in different countries. Secondly EV/EBITDA ratio is less volatile as
items like taxes and interest that can cause a valuation metric to be volatile are not a part of this
ratio. Another important point to be noted is that the EV/EBITDA ratio is not affected by
changes in capital structure.
Price to cash flow ratio (share price divided by cash flow per share) is another useful valuation
metric that is used. The ratio makes use of operating cash flow. This ratio is particularly useful
for valuing those companies that have positive cash flows but are not profitable because of high
amount of non-cash charges like depreciation and amortization. This ratio is useful because of
the fact that cash flows cannot easily be manipulated.
Other market multiple ratios that can be used are P/E (price-earnings) ratio, price to book ratio,
EV/NOPAT ratio and price to sales ratio. These ratios are useful as it helps an investor to
determine the market value of the asset relative to a key statistic that bears a logical relationship
to the market value of the asset in consideration..
What are the respiratory and fermentation pathways in Chlamydia pneu.pdf
What are the respiratory and fermentation pathways in Chlamydia pneumoniae?
Solution
Chlamydophila pneumoniae is, an obligate intracellular bacterium which infects humans and is
one of the major cause of pneumonia. It is gram negative bacterium. As it is an obligate parasite
it is difficult to distinguish between the host and parasite metabolism. There are major
differences in the metabolic pathways in various developmental stages.
C. pneumoniae requires oxygen to survive, but requires environments with lower levels of
oxygen than present in the atmosphere
These bacteria shows metabolism of glucose-6-phosphate to pyruvate via glycolysis, which
allows them to generate ATP via substrate-level phosphorylation by the enzymes
phosphoglycerate kinase and pyruvate kinase.
The TCA cycle is complete in all Chlamydiae thus able to oxidize acetyl-CoA, accompanied by
regeneration of the energy-rich molecules NADH, FADH2 and GTP.
The electron transport chain of environmental chlamydiae is branched. H+ translocating NADH
dehydrogenase, additional terminal oxidases, and cytochrome c oxidases, altogether were
suggested to enable these bacteria to build up a proton gradient more efficiently and potentially
allow them to respond to changing conditions such as fluctuating oxygen tensions.
They also have increased capacities for host-independent production of ATP via oxidative
phosphorylation..
You have a piece of circular DNA that can be cut by the restriction .pdf
You have a piece of circular DNA that can be cut by the restriction nucleases EcoRI, HindIII,
and NotI, as indicated in the Figure Below...
Answer whether the following statement is TRUE or FALSE. If FALSE, explain why.
(a) One piece of DNA will be obtained when this DNA is cut by NotI.
(b) A piece of DNA that cannot be cut by EcoRI will be obtained by cutting this DNA with both
NotI and HindIII.
(c) Two DNA fragments that cannot be cut by HindIII will be obtained when this DNA is cut by
EcoRI and NotI.
(d)Two DNA fragments of unequal size will be created when this DNA is cut by both HindIII
and EcoRI. Notl EcoRI EcoRI HindIII
Solution
1. True
Yes , if the DNA Is cut by only NotI, then only a single piece of DNA would be formed as there
would be a cut in the circular DNA , thus forming a single piece single stranded DNA.
2. False
Even if there is a cut with both NotI and HindIII there would be a piece of DNA between NotI
and ECoRI which will still remain and won\'t be cut.
3. True
Two DNA fragments will be obtained when cut with EcoRI and NotI that cannot be obtained
with HindIII. A fragment between NotI and ECoRI and between the two ECoRI.
4. False
Three unequal pieces of DNA would be formed. First, the HindIII, would cause a nick and the
strand would open and form a single strand. Second, ECoRI would cause two nicks, as it is at
two places, thus two more fragments, one between the two ECoRI and the other between the
ECoRI and HindIII would be formed..
Write a 2-3 page report on any one of the following topics. A support.pdf
Write a 2-3 page report on any one of the following topics. A supporting paper is provided for
each topic. However, the report has to be prepared based on general literature beyond the
supporting paper. The deadline for the submission of the assignment is 14th January, 12 noon,
2017. Your report should have proper references/bibliography. Plagiarism is strictly prohibited!
Topic 1. Impact of invasive species on native species and ecosystems. Supporting paper/s:
Gurevitch, Jessica, and Dianna K. Padilla. \"Are invasive species a major cause of extinctions?-
Trends in Ecology & Evolution 19.9 (2004): 470-474. Molnar, Jennifer L., et al. \"Assessing the
global threat of invasive species to marine biodiversity \" Frontiers in Ecology and the
Environment 6.9 (2008): 485-492. Didham, Raphael K., et al. \"Are invasive species the drivers
of ecological change?.\" Trends in Ecology & Evolution 20.9 (2005): 470-474. strayer, David
L., et a \"understanding the long-term effects of species invasions.\" Trends in ecology &
evolution 21.11 (2006): 645-651.
Solution
Impact of Invasive species on native species and ecosystems
Native species are those originally originated and evolved in the region (Cox, 2004). The native
species adapted the environment and co evolved with other species and the ecosystem was also
evolved accordingly (Cox, 2004). When a foreign/ alien species is introduced to a new
environment by general rule it don’t survive or the new species struggle to adapt the new
environment and struggle to make its place in the ecosystem (Dunlap, 1999; Gonzalez-Browne et
al., 2016). Surprisingly some species better adopt to new environment i.e. the new environment
suits the species more than the original environment (Dunlap, 1999). Few species becomes
invasive and the species population booms in the newly invaded territory (Alexander et al.,
2014). The invasive species becomes overweighed in the natural balance of the native ecosystem
and it inversely impact on the growth of some native species those shares common niches, food,
breeding places, resources or the alien species predates on certain native species (Hickman et al.,
2013; Kovacs et al., 2012). The invading species can also be a new prey for certain native
species which will increase the population of that predating species (Anson and Dickman, 2013;
Barber et al., 2008). Not only the direct relation of prey and predator or competition influences
native species but the introduction of the invading species can also indirectly influence the
abundance of the native species (Mello and Oliveira, 2016). The invasive species also interact
with the new environment and the advantages it gets in the new ecotype also put impact on the
evolution of the invasive species itself as the invasive species evolve becomes different from its
original native state or source species (Zenni et al., 2016). Sometime the invading introduced
species becomes free from its natural pathogen and free from immunological burden.
Why are sex cells (gametes) haploid instead of diploid. Write a brief.pdf
Why are sex cells (gametes) haploid instead of diploid. Write a brief explanation/describe the
what the consequences would be if gametes were diploid instead.
Solution
Sex cells (gametes) are haploid because they are formed by the process of reductional division
called meiosis.
Meiosis is a defined as double division of parent cell which results in the formation of 4 daughter
cells each having half the number of chromsome than that of parent cell. It is the type of cell
division in which chromosome number is reduced from diploid number (2n) to haploid number
(n).
This type of division is important while the formation of gametes as it maintains constant
number of chromsomes generation after generation. If meiosis woulf not occur, gametes with
diploid chromosome number will be produced and when these gametes will fuse to form zygote,
the resulting zygote will have an abnormal ,double the number of chromosome (tetraploid) than
that of the normal number of chromosome found in species. Tetraploidy is very lethal in animals
, therefore meiosis is important in maintaining normal chromosome number..
What role did Warfarin play in the development of pesticide-resistan.pdf
What role did Warfarin play in the development of pesticide-resistant rats?
A. It caused the mutations, which lead to its resistance.
B. It killed off the non-resistant rats.
C. It selected for resistant rats.
D. Both b and c.
Solution
Answer is \"D\". Both b and c.
It kille off the non-resistant rats.
It selected for resistant rats.
Warfarin is a special type of medication used as a blood thinner..
What is the role of health risk appraisal in health promotion and.pdf
What is the role of health risk appraisal in health promotion and disease prevention?
What is the role of health risk appraisal in health promotion and disease prevention?
Solution
The role of health risk appraisal in health promotion and disease prevention is :
Thus, for both health promotion advises and health disease prevention steps requires proper and
detailed appraisal of the present and past medical histories of the cell..
What is Rev protein in the HIV genome and why is it important in the.pdf
What is Rev protein in the HIV genome and why is it important in the HIV life cycle?
Solution
The two most important regulatory factors for HIV (human immuno deficiency virus) gene
expression are, “Tat and Rev.” REV is a phosphoprotein present in the nucleus or nucleolus. It
binds to the RRE and functions to promote the nuclear export, functional activation of RRE
containing viral mRNAs and stabilization. In lentiviruses, it is the most functionally conserved
regulatory protein..
Under what condition would field research be undertaken a. The rese.pdf
Field research would be undertaken if the researcher needs to draw clear conclusions about the population being studied. Direct observation in a natural setting allows the researcher to understand behaviors and make inferences about the larger population based on what is observed.
Sue heads due west from Dry Gulch on U.S. 6.5 at a steady speed of 5.pdf
Sue heads due west from Dry Gulch on U.S. 6.5 at a steady speed of 55 mph (miles per hour)
while Don heads due east from Wet Hollow (which is 830 miles west of Dry Gulch) on the same
hiway at 65 mph but he starts 2 hours after Sue did. How many miles did Don travel before they
met?
Solution
Two hours after Sue started, their distance is 850 - 55*2 = 740 miles.
Then , another T = 740/(55+65) = 740/120 = 37/6 hrs, they met.
And so Don drove 65 * 37/6 = 2405/6 miles before thy met..
What does “LUCA” stand for and what evidence does LUCA provide about.pdf
What does “LUCA” stand for and what evidence does LUCA provide about the possible origins
of life?
Solution
Ans. LUCA stands for ‘Last Universal Common Ancestor’ that evolved around 3.7 billion years
ago. It is the most recent cell (a thermophile) from which all species currently living on Earth are
believed to have evolved. LUCA is believed to have exhibited the features of a present-day cell.
It possibly had DNA as genetic material, DNA-based machinery for the processes of central
dogma, DNA-based genetic code, aqueous cytoplasm surrounded by lipid bilayer and many other
compositions similar to present day prokaryotic cells.
The comparison of 16S rRNA of organisms, currently living or extant, suggest that the three
domains of life (Archaea, Bacteria and Eukarya) have LUCA as their common ancestor. The
cellular organization and functions of these organisms are also similar to LUCA.
Therefore, It presents an evidence that all present day organisms (DNA-based cellular life forms)
evolved from a common ancestor LUCA (DNA-based cellular life form)..
The figure below represents a phylogenetic tree of life. Each color r.pdf
The figure below represents a phylogenetic tree of life. Each color represents a separate domain.
Assign the appropriate name to each of the three domains. Not all the names will be used.
Solution
Blue colour - Bacteria (Gram positive bacteria, green filamentous etc)
green colour - Archaea (Halophile, T.celer etc)
maroon colour - Eukayota (Animal, fungi, plant etc).
Side 4 Knowledge Check 01 A classifed balance sheet commonly include.pdf
Side 4 Knowledge Check 01 A classifed balance sheet commonly includes the tems set forth
below identiry the order in which these tems woutd appear by dragging m n order, into the
appropriate secions of the statement assets Cument tabie Intangible ass Common sck Assets
Equity OB 5 6 8 0 K L
Solution
Assets Liabilities Equity Current assets Current Laibilities Common Stock Long-term
investments Long-term Laibilities Retained Earnings Plant assets Intangible assets.
SG reports net income of $38,000 for 2015. At the beginning of 2015,.pdf
SG reports net income of $38,000 for 2015. At the beginning of 2015, SG had $169,000 in
assets. By 2015, assets had grown to $219,000.Choose Numertor /Choose Denomenator =Return
Assets/=
Solution
Average assets=(Beginning balance+Ending balance)/2
which is equal to
=(169000+219000)/2=$194000Net income/Average assets=Return
Assets38000/194000=19.59%(Approx)..
1. The E. coli phosphotransferase system transports glucose into the.pdf
1. The E. coli phosphotransferase system transports glucose into the cell by:
A. serving as a carrier protein that pumps glucose into the cell in an energy-dependent manner
against its concentration gradient.
B. serving as a carrier protein that pumps glucose into the cell in an energy-independent manner
against its concentration gradient.
C. keeping levels of intracellular glucose very low by efficiently phosphorylating intracellular
glucose in an energy-dependent set of reactions.
D. keeping levels of intracellular glucose very low by efficiently phosphorylating intracellular
glucose in a set of spontaneous reactions.
E. none of the above.
2. The cytoplasmic membrane is a
A. highly selective permeability barrier of the cell.
B. primary structure that maintains the shape of the cell.
C. source of nutrient production.
D. structure which identifies a cell as eukaryotic or prokaryotic.
E. none of the above
3. Light will be __________________ when it interacts with an object that is smaller than the
light\'s wavelength.
A) refracted.
B) reflected
C) absorbed.
D) scattered.
e) transformed
Solution
1. Option C is the correct answer. This process chemically alters the glucose by phosphorylation
in an energy dependent manner. This way glucose enters the cell and is then phosphorylated to
form glucose -6- phosphate. This change maintains the concentration gradient and allows more
glucose to flow into the cell. As glucose -6- phosphate is unable to move out of the cell. Thus
intercellular levels increase and intracellular level remains low..
Read the scenario that you will use for the Individual Projects in ea.pdf
Read the scenario that you will use for the Individual Projects in each week of the course. The
Centers for Medicare and Medicaid Services (CMS) has taken on a more visible role in health
care delivery. Many changes have transpired to improve patient safety along with the
implementation of additional quality metrics, and these changes impact reimbursement rates
Likewise, the Patient Protection and Affordable Care Act has changed the reimbursement fee
structure of Medicare and Medicaid reimbursement for health care services. Other legislation
including the HITECH Act and the Medicare Authorization and CHIP Reactivation Act of 2015
(MACRA) all impact how healthcare organizations receive reimbursement and demonstrate use
of data to improve quality and delivery of patient care Mr. Magone, CEO of Healing Hands
Hospital, has asked you to join the \"Future of Healing Hands Task Force, and your first
assignment is to work with the Hospital Chief Financial Officer, Mr. Johnson, and provide a
summary of the current regulations regarding Medicare reimbursement including how MACR
impact reimbursement if/when Healing Hands coordinates delivery of services by affiliating with
physician practices For this assignment, write a 2-3 page report that you will deliver to Mr.
Magone on how the new CMS initiatives and regulations impact the organization\'s revenue
structure. In your presentation, address the following questions: Why did CMS become more
involved in the reimbursement component of health care? How does CMS\'s involvement impact
the reimbursement model for Healing Hands Hospital and other health care organizations If
CMS reimbursement regulations for Medicare and Medicaid change, does it follow that other
insurance providers change heir policies on reimbursement? What tools can be implemented to
ensure organizations such as Healing Hands Hospital and physician practices are meeting the
policies and procedures set forth by CMS? Identify 3 tools from the CMS Web site that are
helpful in meeting the requirements for Medicare reimbursement set forth by CMS
Solution
Part-a & part-b:
The physician’s work, practice expense, and malpractice, RVU values, CMS (centers for
Medicare and Medicaid services) is required to control overall expenditures in health care
organization. Therefore, CMS become highly involved in the reimbursement component of
health care to patients as per their \"insurance packages\". The CMS\' involvement in “budget
Neutrality” & the reimbursement model at Healing Hand hospital & other health care
organizations is mainly for physician RVU based payments from Medicare & Medicare that can
control its physician costs by adjusting physician payment rates based on “previous periods in a
calendar year” as per federal acts and regulations. The Medicare is going to control physicians
costs according to “medical procedures and medical visits of their record” in a Jan- 1 ending Dec
31. Conversion Factor is main basis to control the physician costs ac.
Q1) Explain the difference between SSL connection and session.Q2) .pdf
Q1) Explain the difference between SSL connection and session.
Q2) What is the advantage of not including the MAC in the scope of packet encryption in SSH
Transport Protocol?
Q3) Show what part of SSL that protects against the following attacks :-
a. Replay Attack: Earlier SSL handshake messages are replayed.
b. Man-in-the-Middle Attack: An attacker interposes during key exchange,
acting as the client to the server and as the server to the client.
c. IP Spoofing: Uses forged IP addresses to fool a host into accepting bogus data.
Solution
Answer :
Q1) Explain the difference between SSL connection and session.
Answer :
Explanation :
These are the main the difference between SSL connection and session..
....................................
Q2) What is the advantage of not including the MAC in the scope of packet encryption in SSH
Transport Protocol?
Answer :
This allows for the message to be authenticated before attempting decryption,which may be more
efficient.
Explanation :
This allows for the message to be authenticated before attempting decryption,which may be more
efficient is the advantage of not including the MAC in the scope of packet encryption in SSH
Transport Protocol
.................S.noConnection Session1.Connection is a live communication channel.Session is a
set of negotiated cryptography parameters.2.A transient peer -to - peer communications
link.Session is an association between client and server.3.Each connection associated with 1 SSL
Session.Session is created by Handshake protocol.4.SSL Connection is a transport that provides
a suitable type of service.Shared by multiple SSL Connections..
Production of extracellular polymeric substances (EPS) enables bacter.pdf
Production of extracellular polymeric substances (EPS) enables bacteria to form biofilms, make
endospores. grow in cold climates such as the Arctic. communicate with other cells. undergo
DNA mediated transformation.
Solution
Production of extracellular polymeric substances (EPS) enables bacteria to form biofilms
b.make endospores
c.grow in cold climates such as the Arctic
d. Communicate with other cells
e.Undergo DNA mediated transformation
Explaination:
EPSs establish the structural and functional integrity of the biofilms, and these are considered as
the fundamental components that determines the physiochemical properties of a biofilm. So
answer is option A.
More Related Content
More from ashokarians
Turner syndrome is caused by monosomy of the sex chromosomes XO. .pdf
Turner syndrome is caused by monosomy of the sex chromosomes XO. Why is there no
corresponding condition for monosomy of the sex chromosome YO
Turner syndrome is caused by monosomy of the sex chromosomes XO. Why is there no
corresponding condition for monosomy of the sex chromosome YO
Solution
Nondisjunction leading to YO cells are non viable.At least one X chromosome is needed for
viability.
YO condition is lethal hence there are no corresponding conditions for monosomy of the sex
chromosome YO..
why are we seeing more evidence of widespread etchical dilemmas with.pdf
why are we seeing more evidence of widespread etchical dilemmas within organization?
Solution
Social Responsibility and caring and trust worthiness has been started to decrease, the reason is
thinking about more profit with less effort.
Lack of internal cooperation and Misunderstanding between the workers, it was raised..
What do evebitda ratio and price to cash flow ratio show us .pdf
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
What do ev/ebitda ratio and price to cash flow ratio show us and how do they compare?
Evaluate the usefulness of these and other market multiple ratios in company evaluations
Solution
EV/EBITDA ratio (enterprise value divided by earnings before interest, tax, depreciation and
amortization) is an important valuation multiple and the reason why it is so popular is that the
ratio makes adjustments so that an investor can identify the real cash flow issue associated with a
company. The ratio shows the real value of a business and hence can be used to easily compare
companies that operate in different countries. Secondly EV/EBITDA ratio is less volatile as
items like taxes and interest that can cause a valuation metric to be volatile are not a part of this
ratio. Another important point to be noted is that the EV/EBITDA ratio is not affected by
changes in capital structure.
Price to cash flow ratio (share price divided by cash flow per share) is another useful valuation
metric that is used. The ratio makes use of operating cash flow. This ratio is particularly useful
for valuing those companies that have positive cash flows but are not profitable because of high
amount of non-cash charges like depreciation and amortization. This ratio is useful because of
the fact that cash flows cannot easily be manipulated.
Other market multiple ratios that can be used are P/E (price-earnings) ratio, price to book ratio,
EV/NOPAT ratio and price to sales ratio. These ratios are useful as it helps an investor to
determine the market value of the asset relative to a key statistic that bears a logical relationship
to the market value of the asset in consideration..
What are the respiratory and fermentation pathways in Chlamydia pneu.pdf
What are the respiratory and fermentation pathways in Chlamydia pneumoniae?
Solution
Chlamydophila pneumoniae is, an obligate intracellular bacterium which infects humans and is
one of the major cause of pneumonia. It is gram negative bacterium. As it is an obligate parasite
it is difficult to distinguish between the host and parasite metabolism. There are major
differences in the metabolic pathways in various developmental stages.
C. pneumoniae requires oxygen to survive, but requires environments with lower levels of
oxygen than present in the atmosphere
These bacteria shows metabolism of glucose-6-phosphate to pyruvate via glycolysis, which
allows them to generate ATP via substrate-level phosphorylation by the enzymes
phosphoglycerate kinase and pyruvate kinase.
The TCA cycle is complete in all Chlamydiae thus able to oxidize acetyl-CoA, accompanied by
regeneration of the energy-rich molecules NADH, FADH2 and GTP.
The electron transport chain of environmental chlamydiae is branched. H+ translocating NADH
dehydrogenase, additional terminal oxidases, and cytochrome c oxidases, altogether were
suggested to enable these bacteria to build up a proton gradient more efficiently and potentially
allow them to respond to changing conditions such as fluctuating oxygen tensions.
They also have increased capacities for host-independent production of ATP via oxidative
phosphorylation..
You have a piece of circular DNA that can be cut by the restriction .pdf
You have a piece of circular DNA that can be cut by the restriction nucleases EcoRI, HindIII,
and NotI, as indicated in the Figure Below...
Answer whether the following statement is TRUE or FALSE. If FALSE, explain why.
(a) One piece of DNA will be obtained when this DNA is cut by NotI.
(b) A piece of DNA that cannot be cut by EcoRI will be obtained by cutting this DNA with both
NotI and HindIII.
(c) Two DNA fragments that cannot be cut by HindIII will be obtained when this DNA is cut by
EcoRI and NotI.
(d)Two DNA fragments of unequal size will be created when this DNA is cut by both HindIII
and EcoRI. Notl EcoRI EcoRI HindIII
Solution
1. True
Yes , if the DNA Is cut by only NotI, then only a single piece of DNA would be formed as there
would be a cut in the circular DNA , thus forming a single piece single stranded DNA.
2. False
Even if there is a cut with both NotI and HindIII there would be a piece of DNA between NotI
and ECoRI which will still remain and won\'t be cut.
3. True
Two DNA fragments will be obtained when cut with EcoRI and NotI that cannot be obtained
with HindIII. A fragment between NotI and ECoRI and between the two ECoRI.
4. False
Three unequal pieces of DNA would be formed. First, the HindIII, would cause a nick and the
strand would open and form a single strand. Second, ECoRI would cause two nicks, as it is at
two places, thus two more fragments, one between the two ECoRI and the other between the
ECoRI and HindIII would be formed..
Write a 2-3 page report on any one of the following topics. A support.pdf
Write a 2-3 page report on any one of the following topics. A supporting paper is provided for
each topic. However, the report has to be prepared based on general literature beyond the
supporting paper. The deadline for the submission of the assignment is 14th January, 12 noon,
2017. Your report should have proper references/bibliography. Plagiarism is strictly prohibited!
Topic 1. Impact of invasive species on native species and ecosystems. Supporting paper/s:
Gurevitch, Jessica, and Dianna K. Padilla. \"Are invasive species a major cause of extinctions?-
Trends in Ecology & Evolution 19.9 (2004): 470-474. Molnar, Jennifer L., et al. \"Assessing the
global threat of invasive species to marine biodiversity \" Frontiers in Ecology and the
Environment 6.9 (2008): 485-492. Didham, Raphael K., et al. \"Are invasive species the drivers
of ecological change?.\" Trends in Ecology & Evolution 20.9 (2005): 470-474. strayer, David
L., et a \"understanding the long-term effects of species invasions.\" Trends in ecology &
evolution 21.11 (2006): 645-651.
Solution
Impact of Invasive species on native species and ecosystems
Native species are those originally originated and evolved in the region (Cox, 2004). The native
species adapted the environment and co evolved with other species and the ecosystem was also
evolved accordingly (Cox, 2004). When a foreign/ alien species is introduced to a new
environment by general rule it don’t survive or the new species struggle to adapt the new
environment and struggle to make its place in the ecosystem (Dunlap, 1999; Gonzalez-Browne et
al., 2016). Surprisingly some species better adopt to new environment i.e. the new environment
suits the species more than the original environment (Dunlap, 1999). Few species becomes
invasive and the species population booms in the newly invaded territory (Alexander et al.,
2014). The invasive species becomes overweighed in the natural balance of the native ecosystem
and it inversely impact on the growth of some native species those shares common niches, food,
breeding places, resources or the alien species predates on certain native species (Hickman et al.,
2013; Kovacs et al., 2012). The invading species can also be a new prey for certain native
species which will increase the population of that predating species (Anson and Dickman, 2013;
Barber et al., 2008). Not only the direct relation of prey and predator or competition influences
native species but the introduction of the invading species can also indirectly influence the
abundance of the native species (Mello and Oliveira, 2016). The invasive species also interact
with the new environment and the advantages it gets in the new ecotype also put impact on the
evolution of the invasive species itself as the invasive species evolve becomes different from its
original native state or source species (Zenni et al., 2016). Sometime the invading introduced
species becomes free from its natural pathogen and free from immunological burden.
Why are sex cells (gametes) haploid instead of diploid. Write a brief.pdf
Why are sex cells (gametes) haploid instead of diploid. Write a brief explanation/describe the
what the consequences would be if gametes were diploid instead.
Solution
Sex cells (gametes) are haploid because they are formed by the process of reductional division
called meiosis.
Meiosis is a defined as double division of parent cell which results in the formation of 4 daughter
cells each having half the number of chromsome than that of parent cell. It is the type of cell
division in which chromosome number is reduced from diploid number (2n) to haploid number
(n).
This type of division is important while the formation of gametes as it maintains constant
number of chromsomes generation after generation. If meiosis woulf not occur, gametes with
diploid chromosome number will be produced and when these gametes will fuse to form zygote,
the resulting zygote will have an abnormal ,double the number of chromosome (tetraploid) than
that of the normal number of chromosome found in species. Tetraploidy is very lethal in animals
, therefore meiosis is important in maintaining normal chromosome number..
What role did Warfarin play in the development of pesticide-resistan.pdf
What role did Warfarin play in the development of pesticide-resistant rats?
A. It caused the mutations, which lead to its resistance.
B. It killed off the non-resistant rats.
C. It selected for resistant rats.
D. Both b and c.
Solution
Answer is \"D\". Both b and c.
It kille off the non-resistant rats.
It selected for resistant rats.
Warfarin is a special type of medication used as a blood thinner..
What is the role of health risk appraisal in health promotion and.pdf
What is the role of health risk appraisal in health promotion and disease prevention?
What is the role of health risk appraisal in health promotion and disease prevention?
Solution
The role of health risk appraisal in health promotion and disease prevention is :
Thus, for both health promotion advises and health disease prevention steps requires proper and
detailed appraisal of the present and past medical histories of the cell..
What is Rev protein in the HIV genome and why is it important in the.pdf
What is Rev protein in the HIV genome and why is it important in the HIV life cycle?
Solution
The two most important regulatory factors for HIV (human immuno deficiency virus) gene
expression are, “Tat and Rev.” REV is a phosphoprotein present in the nucleus or nucleolus. It
binds to the RRE and functions to promote the nuclear export, functional activation of RRE
containing viral mRNAs and stabilization. In lentiviruses, it is the most functionally conserved
regulatory protein..
Under what condition would field research be undertaken a. The rese.pdf
Field research would be undertaken if the researcher needs to draw clear conclusions about the population being studied. Direct observation in a natural setting allows the researcher to understand behaviors and make inferences about the larger population based on what is observed.
Sue heads due west from Dry Gulch on U.S. 6.5 at a steady speed of 5.pdf
Sue heads due west from Dry Gulch on U.S. 6.5 at a steady speed of 55 mph (miles per hour)
while Don heads due east from Wet Hollow (which is 830 miles west of Dry Gulch) on the same
hiway at 65 mph but he starts 2 hours after Sue did. How many miles did Don travel before they
met?
Solution
Two hours after Sue started, their distance is 850 - 55*2 = 740 miles.
Then , another T = 740/(55+65) = 740/120 = 37/6 hrs, they met.
And so Don drove 65 * 37/6 = 2405/6 miles before thy met..
What does “LUCA” stand for and what evidence does LUCA provide about.pdf
What does “LUCA” stand for and what evidence does LUCA provide about the possible origins
of life?
Solution
Ans. LUCA stands for ‘Last Universal Common Ancestor’ that evolved around 3.7 billion years
ago. It is the most recent cell (a thermophile) from which all species currently living on Earth are
believed to have evolved. LUCA is believed to have exhibited the features of a present-day cell.
It possibly had DNA as genetic material, DNA-based machinery for the processes of central
dogma, DNA-based genetic code, aqueous cytoplasm surrounded by lipid bilayer and many other
compositions similar to present day prokaryotic cells.
The comparison of 16S rRNA of organisms, currently living or extant, suggest that the three
domains of life (Archaea, Bacteria and Eukarya) have LUCA as their common ancestor. The
cellular organization and functions of these organisms are also similar to LUCA.
Therefore, It presents an evidence that all present day organisms (DNA-based cellular life forms)
evolved from a common ancestor LUCA (DNA-based cellular life form)..
The figure below represents a phylogenetic tree of life. Each color r.pdf
The figure below represents a phylogenetic tree of life. Each color represents a separate domain.
Assign the appropriate name to each of the three domains. Not all the names will be used.
Solution
Blue colour - Bacteria (Gram positive bacteria, green filamentous etc)
green colour - Archaea (Halophile, T.celer etc)
maroon colour - Eukayota (Animal, fungi, plant etc).
Side 4 Knowledge Check 01 A classifed balance sheet commonly include.pdf
Side 4 Knowledge Check 01 A classifed balance sheet commonly includes the tems set forth
below identiry the order in which these tems woutd appear by dragging m n order, into the
appropriate secions of the statement assets Cument tabie Intangible ass Common sck Assets
Equity OB 5 6 8 0 K L
Solution
Assets Liabilities Equity Current assets Current Laibilities Common Stock Long-term
investments Long-term Laibilities Retained Earnings Plant assets Intangible assets.
SG reports net income of $38,000 for 2015. At the beginning of 2015,.pdf
SG reports net income of $38,000 for 2015. At the beginning of 2015, SG had $169,000 in
assets. By 2015, assets had grown to $219,000.Choose Numertor /Choose Denomenator =Return
Assets/=
Solution
Average assets=(Beginning balance+Ending balance)/2
which is equal to
=(169000+219000)/2=$194000Net income/Average assets=Return
Assets38000/194000=19.59%(Approx)..
1. The E. coli phosphotransferase system transports glucose into the.pdf
1. The E. coli phosphotransferase system transports glucose into the cell by:
A. serving as a carrier protein that pumps glucose into the cell in an energy-dependent manner
against its concentration gradient.
B. serving as a carrier protein that pumps glucose into the cell in an energy-independent manner
against its concentration gradient.
C. keeping levels of intracellular glucose very low by efficiently phosphorylating intracellular
glucose in an energy-dependent set of reactions.
D. keeping levels of intracellular glucose very low by efficiently phosphorylating intracellular
glucose in a set of spontaneous reactions.
E. none of the above.
2. The cytoplasmic membrane is a
A. highly selective permeability barrier of the cell.
B. primary structure that maintains the shape of the cell.
C. source of nutrient production.
D. structure which identifies a cell as eukaryotic or prokaryotic.
E. none of the above
3. Light will be __________________ when it interacts with an object that is smaller than the
light\'s wavelength.
A) refracted.
B) reflected
C) absorbed.
D) scattered.
e) transformed
Solution
1. Option C is the correct answer. This process chemically alters the glucose by phosphorylation
in an energy dependent manner. This way glucose enters the cell and is then phosphorylated to
form glucose -6- phosphate. This change maintains the concentration gradient and allows more
glucose to flow into the cell. As glucose -6- phosphate is unable to move out of the cell. Thus
intercellular levels increase and intracellular level remains low..
Read the scenario that you will use for the Individual Projects in ea.pdf
Read the scenario that you will use for the Individual Projects in each week of the course. The
Centers for Medicare and Medicaid Services (CMS) has taken on a more visible role in health
care delivery. Many changes have transpired to improve patient safety along with the
implementation of additional quality metrics, and these changes impact reimbursement rates
Likewise, the Patient Protection and Affordable Care Act has changed the reimbursement fee
structure of Medicare and Medicaid reimbursement for health care services. Other legislation
including the HITECH Act and the Medicare Authorization and CHIP Reactivation Act of 2015
(MACRA) all impact how healthcare organizations receive reimbursement and demonstrate use
of data to improve quality and delivery of patient care Mr. Magone, CEO of Healing Hands
Hospital, has asked you to join the \"Future of Healing Hands Task Force, and your first
assignment is to work with the Hospital Chief Financial Officer, Mr. Johnson, and provide a
summary of the current regulations regarding Medicare reimbursement including how MACR
impact reimbursement if/when Healing Hands coordinates delivery of services by affiliating with
physician practices For this assignment, write a 2-3 page report that you will deliver to Mr.
Magone on how the new CMS initiatives and regulations impact the organization\'s revenue
structure. In your presentation, address the following questions: Why did CMS become more
involved in the reimbursement component of health care? How does CMS\'s involvement impact
the reimbursement model for Healing Hands Hospital and other health care organizations If
CMS reimbursement regulations for Medicare and Medicaid change, does it follow that other
insurance providers change heir policies on reimbursement? What tools can be implemented to
ensure organizations such as Healing Hands Hospital and physician practices are meeting the
policies and procedures set forth by CMS? Identify 3 tools from the CMS Web site that are
helpful in meeting the requirements for Medicare reimbursement set forth by CMS
Solution
Part-a & part-b:
The physician’s work, practice expense, and malpractice, RVU values, CMS (centers for
Medicare and Medicaid services) is required to control overall expenditures in health care
organization. Therefore, CMS become highly involved in the reimbursement component of
health care to patients as per their \"insurance packages\". The CMS\' involvement in “budget
Neutrality” & the reimbursement model at Healing Hand hospital & other health care
organizations is mainly for physician RVU based payments from Medicare & Medicare that can
control its physician costs by adjusting physician payment rates based on “previous periods in a
calendar year” as per federal acts and regulations. The Medicare is going to control physicians
costs according to “medical procedures and medical visits of their record” in a Jan- 1 ending Dec
31. Conversion Factor is main basis to control the physician costs ac.
Q1) Explain the difference between SSL connection and session.Q2) .pdf
Q1) Explain the difference between SSL connection and session.
Q2) What is the advantage of not including the MAC in the scope of packet encryption in SSH
Transport Protocol?
Q3) Show what part of SSL that protects against the following attacks :-
a. Replay Attack: Earlier SSL handshake messages are replayed.
b. Man-in-the-Middle Attack: An attacker interposes during key exchange,
acting as the client to the server and as the server to the client.
c. IP Spoofing: Uses forged IP addresses to fool a host into accepting bogus data.
Solution
Answer :
Q1) Explain the difference between SSL connection and session.
Answer :
Explanation :
These are the main the difference between SSL connection and session..
....................................
Q2) What is the advantage of not including the MAC in the scope of packet encryption in SSH
Transport Protocol?
Answer :
This allows for the message to be authenticated before attempting decryption,which may be more
efficient.
Explanation :
This allows for the message to be authenticated before attempting decryption,which may be more
efficient is the advantage of not including the MAC in the scope of packet encryption in SSH
Transport Protocol
.................S.noConnection Session1.Connection is a live communication channel.Session is a
set of negotiated cryptography parameters.2.A transient peer -to - peer communications
link.Session is an association between client and server.3.Each connection associated with 1 SSL
Session.Session is created by Handshake protocol.4.SSL Connection is a transport that provides
a suitable type of service.Shared by multiple SSL Connections..
Production of extracellular polymeric substances (EPS) enables bacter.pdf
Production of extracellular polymeric substances (EPS) enables bacteria to form biofilms, make
endospores. grow in cold climates such as the Arctic. communicate with other cells. undergo
DNA mediated transformation.
Solution
Production of extracellular polymeric substances (EPS) enables bacteria to form biofilms
b.make endospores
c.grow in cold climates such as the Arctic
d. Communicate with other cells
e.Undergo DNA mediated transformation
Explaination:
EPSs establish the structural and functional integrity of the biofilms, and these are considered as
the fundamental components that determines the physiochemical properties of a biofilm. So
answer is option A.
More from ashokarians (20)
Turner syndrome is caused by monosomy of the sex chromosomes XO. .pdf
Turner syndrome is caused by monosomy of the sex chromosomes XO. .pdf
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