Side 4 Knowledge Check 01 A classifed balance sheet commonly include.pdf

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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.

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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..

Turner syndrome is caused by monosomy of the sex chromosomes XO. .pdf

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why are we seeing more evidence of widespread etchical dilemmas with.pdf

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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 do evebitda ratio and price to cash flow ratio show us .pdf

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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..

What are the respiratory and fermentation pathways in Chlamydia pneu.pdf

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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..

You have a piece of circular DNA that can be cut by the restriction .pdf

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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.

Write a 2-3 page report on any one of the following topics. A support.pdf

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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..

Why are sex cells (gametes) haploid instead of diploid. Write a brief.pdf

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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 role did Warfarin play in the development of pesticide-resistan.pdf

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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 the role of health risk appraisal in health promotion and.pdf

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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..

What is Rev protein in the HIV genome and why is it important in the.pdf

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Under what condition would field research be undertaken? a. The researcher needs direct control over the variables under study. b. The researcher needs to draw clear conclusions about the population. c. The researcher needs to show rigorous testing of causal hypotheses. d. The researcher needs to understand behavior in a natural setting. Solution b. The researcher needs to draw clear conclusions about the population..

Under what condition would field research be undertaken a. The rese.pdf

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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..

Sue heads due west from Dry Gulch on U.S. 6.5 at a steady speed of 5.pdf

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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)..

What does “LUCA” stand for and what evidence does LUCA provide about.pdf

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The figure below represents a phylogenetic tree of life. Each color r.pdf

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SG reports net income of $38,000 for 2015. At the beginning of 2015,.pdf

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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..

1. The E. coli phosphotransferase system transports glucose into the.pdf

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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.

Read the scenario that you will use for the Individual Projects in ea.pdf

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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..

Q1) Explain the difference between SSL connection and session.Q2) .pdf

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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.

Production of extracellular polymeric substances (EPS) enables bacter.pdf

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Can people who get the 4CMenB vaccine still carry meningococcal B bacteria? Solution 4CMenB (recombinant multicomponent meningococcal B) vaccine will be given to prevent meningitis which is caused by meningococcus and this induces the production of specific bactericidal antibodies again the Men B strain. This meningococcus will be found in throat or nose, but it will not cause any infection in some people, but will induce infection if patient\'s immune system will become weak. 4CMen B (novel multicomponent meningococcal B vaccine) will provide protection against the meningococcal serogroup B. So people got vaccinated with 4CMenB vaccine will not carry meningococcal B bacteria.

Can people who get the 4CMenB vaccine still carry meningococcal B bac.pdf

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