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2:1 Solution 2:1.

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Quick ratio = (Cash + accounts receivable)Accounts payableFor 200.pdf

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Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she require treatment? Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibiotics such as penicillin, corticosteroids and diuretics are to be given). The infection is strep throat and it is caused by streptococcal bacteria (group B). These bacteria infect the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body aches. Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Apart from antibiotics, the major future intervention to treat strep throat is to implement vaccination to boost innate immune levels in the infant because antibiotics such as macrolides have potential inhibitor effects on the growth of the useful gutfolra in infants & toddlers. Probiotics, gicing more Vitamin D & C and natural remedies are major future interventions to avoid strep throat in children. Immunity stimulation is specific future intervention to combat strep throat in infants as vaccination promotes immune factors to reduce the bacteria that cause inflammation of the blood vessels near throat. The infection is strep throat and streptococcal bacteria (group B) cause it. These bacteria infect the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body aches. In a Gram stain of sputum specimen of beta hemolytic streptococci, \"squamous epithelial cell type\" of body cell provides an indication that the sputum specimen represents material from an active infection. Other cells are present in the sputum during the active infection are polymorphonuclear leukocytes (PMNs) Ques-2) What is the significance of facial swelling? List the possible causes of facial swelling? 1). Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibiotics such as penicillin, corticosteroids and diuretics are to be given). 2). The bacteria causes inflammation of the blood vessels supplying glomeruli, which impairs the glomerular filtration. Decreased urine filtration causes edema (swelling of body organs). Ques-3) What are the possible causes of dark urine? Dark urine is mainly due to \"presence of RBC\" in the blood. Urine was positive for 3+ protein and for3+ blood by dipstick. The examination of urinary sediment revealed numerous RBCs, many of which were dismorphic. Red blood cell casts were also present. Solution Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she require treatment? Throat infections caused by streptococcal bacteria may cause \"post-streptococcal glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms (antibioti.

Ques-1) What is the most likely cause of a throat infection in a 6-y.pdf

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Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So body’s maximum cells like somatic cells undergo mitosis rather than meiosis. Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this section by multiple sites of duplication. Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. These are non-identical sister chromatids. Part4. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal. Part5. During Prophase I homologous chromosomes physically pair up. This phase includes paring up of chromosomes and homologous recombination. Solution Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So body’s maximum cells like somatic cells undergo mitosis rather than meiosis. Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this section by multiple sites of duplication. Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division. These are non-identical sister chromatids. Part4. The Anaphase II is responsible for the separation of sister chromatids to form chromosomes. Sister chromatids are chromosomes of either maternal or paternal. Part5. During Prophase I homologous chromosomes physically pair up. This phase includes paring up of chromosomes and homologous recombination..

Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf

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Performance Ratios Market Value Added (MVA): MVA=(Company’s market value) minus(Invested capital) Market to book=Market value of shares/Book value of shares Data on market value are not available. Hence the above performance ratios cannot be calculated Profitability Ratios Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity) Dividend information is not available Return on capital(ROC)=(Net Income)/Average (Debt+Equity) Return on asset(ROA)=Net Income/(Average total asset) Return on Equity(ROE)=Net Income/(Average Equity) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 12444.5 F Debt+Equity 20993 21574 21283.5 G Total asset 25707 25509 25608 H=C/F Return on capital(ROC) 0.241454648 I=C/G Return on asset(ROA) 0.200679475 J=C/E Return on Equity(ROE) 0.412953514 Return on capital(ROC) 24.15% Return on asset(ROA) 20.07% Return on Equity(ROE) 41.30% Efficiency Ratios Profit margin=Net Profit/Sales Operating Profit margin=Operating Profit/Sales Asset turnover=Sales/Average Total Assets Inventory turnover(cogs)=Cost of goods sold/Average total inventory Days in inventory=365/inventory turnover(cogs) Inventory turnover is also calculated from sales: Inventory turnover (Sales)=Sales/average total inventory Days sales in inventory=365/inventory turnover (sales) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Operating Income 6946 E Sales 30274 F Total assets 25707 25509 25608 G Cost of goods sold 15383 H Total inventory 3518 3706 3612 I=C/E Profit margin 0.16974962 J=D/E Operating Profit margin 0.229437801 K=E/F Asset turnover 1.182208685 L=G/H Inventory turnover(cogs) 4.258859358 M=365/L Days in inventory 85.70369889 N=E/H Inventory turnover (Sales) 8.381506091 P=365/N Days sales in inventory 43.54825923 Profit margin 16.97% Operating Profit margin 22.94% Asset turnover 1.18220868 Inventory turnover(cogs) 4.25885936 Days in inventory 85.7036989 Days Inventory turnover (Sales) 8.38150609 Days sales in inventory 43.5482592 Days A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 12444.5 F Debt+Equity 20993 21574 21283.5 G Total asset 25707 25509 25608 H=C/F Return on capital(ROC) 0.241454648 I=C/G Return on asset(ROA) 0.200679475 J=C/E Return on Equity(ROE) 0.412953514 Solution Performance Ratios Market Value Added (MVA): MVA=(Company’s market value) minus(Invested capital) Market to book=Market value of shares/Book value of shares Data on market value are not available. Hence the above performance ratios cannot be calculated Profitability Ratios Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity) Dividend information is not available Return on capital(ROC)=(Net Income)/Average (Debt+Equity) Return on asset(ROA)=Net Income/(Average total asset) Return on Equity(ROE)=Net Income/(Average Equity) A B (A+B)/2 Dec 31 2015 Dec 31 2014 2015Average C Net income 5139 D Debt 9246 8432 8839 E Equity 11747 13142 124.

Performance RatiosMarket Value Added (MVA)MVA=(Company’s market.pdf

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package employeeType.employee; public class Employee { private String firstName; private String lastName; private char middleInitial; private boolean fulltime; private char gender; private int employeeNum; public Employee(String firstName, String lastName, char middleInitial, char gender, int employeeNum, boolean fulltime) { super(); this.firstName = firstName; this.lastName = lastName; this.middleInitial = middleInitial; this.fulltime = fulltime; this.gender = gender; this.employeeNum = employeeNum; } public int getEmployeeNumber() { return this.employeeNum; } public void setEmployeeNumber(int empNum) { this.employeeNum = empNum; } public String getFirstName() { return firstName; } String getLastName() { return lastName; } char getMiddleInitial() { return middleInitial; } public char getGender() { return gender; } public void setFirstName(String fn) { this.firstName = fn; } public void setLastName(String ln) { this.lastName = ln; } public void setMiddleI(char m) { this.middleInitial = m; } public void setGender(char g) { this.gender = g; } public boolean equals(Object obj) { // TODO Auto-generated method stub if (obj instanceof Employee) { Employee employee = (Employee) obj; if (employee.getEmployeeNumber() == this.getEmployeeNumber()) return true; else return false; } else return false; } public String toString() { // TODO Auto-generated method stub return getEmployeeNumber() + \"\ \" + getFirstName() + \" \" + getLastName() + \"\ Gender:\" + getGender() + \"\ Status:\" + ((fulltime) ? \"Full Time\" : \"Part Time\"); } } package employeeType.subTypes; import java.text.DecimalFormat; import employeeType.employee.Employee; public class HourlyEmployee extends Employee { private double wage; private double hoursWorked; public HourlyEmployee(String firstName, String lastName, char middleInitial, char gender, int employeeNum, boolean fulltime, double wage) { super(firstName, lastName, middleInitial, gender, employeeNum, fulltime); // TODO Auto-generated constructor stub this.wage = wage; this.hoursWorked = 0.0d; } public void increaseHours(double hours) { if (hours > 0) this.hoursWorked += hours; else System.out.println(\"Error! Invalid hours\"); } public String toString() { // TODO Auto-generated method stub return super.toString() + \"\ Wage: \" + wage + \"\ Hours Worked: \" + hoursWorked; } public double calculateWeeklyPay() { double pay; if (hoursWorked > 40) { pay = (40 * wage) + (2 * (hoursWorked - 40) * wage); } else { pay = wage * hoursWorked; } return pay; } public void annualRaise() { DecimalFormat decimalFormat = new DecimalFormat(\"#.##\"); wage = Double.valueOf(decimalFormat.format(wage * 0.05)); } public double holidayBonus() { return 40 * wage; } public void resetWeek() { this.hoursWorked = 0; } } package employeeType.subTypes; import java.text.DecimalFormat; import employeeType.employee.Employee; public class SalaryEmployee extends Employee { double salary; public SalaryEmployee(String firstName, String lastName, char middleInitia.

package employeeType.employee;public class Employee { private .pdf

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P = 212option ASolutionP = 212option A.pdf

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OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf

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Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf

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NH4NO3 = N2O + 2 H2OMoles of NH4NO3 = 12 x moles of H2O= 12 x .pdf

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No. of shares outstanding = Total assets x weight of equity / price per share = $12,020,000 x50% / $8 = 751250 Net Income = Total assets x Return on assets = $12,020,000 x 15.10% = 1,815,020 Earnings per share = Net Income/ No. of shares outstanding = 1,815,020 / 751250 = 2.416 Solution No. of shares outstanding = Total assets x weight of equity / price per share = $12,020,000 x50% / $8 = 751250 Net Income = Total assets x Return on assets = $12,020,000 x 15.10% = 1,815,020 Earnings per share = Net Income/ No. of shares outstanding = 1,815,020 / 751250 = 2.416.

No. of shares outstanding = Total assets x weight of equity price .pdf

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molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf

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import java.util.Scanner;public class Main { public static in.pdf

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hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf

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O2 is simply oxygen gas. Gases are usually made .pdf

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CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage virus or plasmid. Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the CRISPR/Cpf1 system of Francisella novicida. Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent motif and requiring only a CRISPR RNA (crRNA) for successful targeting. A recent study says that this system has the ability to regulate endogenous genes, responsible for pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response in human host. This immune system in bacteria play role in the regulation of those genes, which are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR immune system also helps bacterium to bypass the human immune system. The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the complex at the correct spot on the double helix to cleave target DNA. CRISPR/Cpf1 systems activity has three stages: 1. Adaptation: Cas1 and Cas2 proteins facilitate the adaptation of small fragments of DNA into the CRISPR array. 2. Formation of crRNAs: processing of pre-cr-RNAs producing of mature crRNAs to guide the Cas protein. 3. Interference: the Cpf1 is bound to a crRNA to form a binary complex to identify and cleave a target DNA sequence. Solution CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage virus or plasmid. Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the CRISPR/Cpf1 system of Francisella novicida. Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent motif and requiring only a CRISPR RNA (crRNA) for successful targeting. A recent study says that this system has the ability to regulate endogenous genes, responsible for pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response in human host. This immune system in bacteria play role in the regulation of those genes, which are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR immune system also helps bacterium to bypass the human immune system. The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the complex at the correct spot on the double helix to cleave target DNA. CRISP.

CRISPR are segments of prokaryotic DNA containing short repetitions .pdf

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C. Router(config-if)#ip access-group 110 in is the right answer.ip.pdf

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C is falseSolutionC is false.pdf

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B). 12SolutionB). 12.pdf

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Answer: Note: LinkedList.cpp is written and driver program main.cpp is also given. //ListInterface.h #ifndef _LIST_INTERFACE #define _LIST_INTERFACE template class ListInterface { public: virtual bool isEmpty() const = 0; virtual int getLength() const = 0; virtual bool insert(int newPosition, const ItemType& newEntry) = 0; virtual bool remove(int position) = 0; virtual void clear() = 0; virtual ItemType getEntry(int position) const = 0; virtual void replace(int position, const ItemType& newEntry) = 0; }; // end ListInterface #endif //Node.h #ifndef NODE_ #define NODE_ template class Node { private: ItemType item; // A data item Node* next; // Pointer to next node public: Node(); Node(const ItemType& anItem); Node(const ItemType& anItem, Node* nextNodePtr); void setItem(const ItemType& anItem); void setNext(Node* nextNodePtr); ItemType getItem() const ; Node* getNext() const ; }; // end Node #endif //Node.cpp #include \"Node.h\" template Node::Node() : next(nullptr) { } // end default constructor template Node::Node(const ItemType& anItem) : item(anItem), next(nullptr) { } // end constructor template Node::Node(const ItemType& anItem, Node* nextNodePtr) : item(anItem), next(nextNodePtr) { } // end constructor template void Node::setItem(const ItemType& anItem) { item = anItem; } // end setItem template void Node::setNext(Node* nextNodePtr) { next = nextNodePtr; } // end setNext template ItemType Node::getItem() const { return item; } // end getItem template Node* Node::getNext() const { return next; } // end getNext //PrecondViolatedExcept.h #ifndef PRECOND_VIOLATED_EXCEPT_ #define PRECOND_VIOLATED_EXCEPT_ #include #include class PrecondViolatedExcept : public std::logic_error { public: PrecondViolatedExcept(const std::string& message = \"\"); }; // end PrecondViolatedExcept #endif //PrecondViolatedExcept.cpp #include \"PrecondViolatedExcept.h\" PrecondViolatedExcept::PrecondViolatedExcept(const std::string& message) : std::logic_error(\"Precondition Violated Exception: \" + message) { } // end constructor //LinkedList.h #ifndef LINKED_LIST_ #define LINKED_LIST_ #include \"ListInterface.h\" #include \"Node.h\" #include #include \"PrecondViolatedExcept.h\" template class LinkedList : public ListInterface { private: Node* headPtr; // Pointer to first node in the chain; // (contains the first entry in the list) int itemCount; // Current count of list items // Locates a specified node in this linked list. Node* getNodeAt(int position) const; public: LinkedList(); LinkedList(const LinkedList& aList); virtual ~LinkedList(); bool isEmpty() const; int getLength() const; bool insert(int newPosition, const ItemType& newEntry); bool remove(int position); void clear(); ItemType getEntry(int position) const; void replace(int position, const ItemType& newEntry); }; // end LinkedList #endif //LinkedList.cpp #include \"LinkedList.h\" #include #include #include #include //Default constructor template LinkedList::LinkedList() : headPtr(nullptr), itemCount(0) { .

AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf

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Answer: The nucleus is separated from the cytoplasm by a phospholipid bilayer known as nuclear envelope. Transport of macromolecules across the nuclear membrane takes place through specialized nuclear pores. Proteins found in the nucleus are synthesized in the cytoplasm and imported into the nucleus through nuclear pore complexes. Such proteins contain a nuclear- localization signal (NLS) that directs their selective transport into the nucleus.4 proteins are required for nuclear localization- importin , importin , NTF-2 & Ran. Answer-A: It has been previously said that nuclear localization signal(NLS) is located within the amino acid sequence of a protein which needs to be localized in the nucleus. Importins actually identify this signal within the protein and interacts with it and facilitates the transfer of nuclear protein from cytosol to nucleus. If the NLS is not present and has already been deleted, the importins won\'t be able to identify the protein and transport them into the nucleus. In this present case, deletion mutation of the protein has been performed to identify the part of the amino acid sequence acting as NLS. In addition leptomycin B has also been added which prevents nuclear export and is not relevant with respect to nuclear localization and NLS. NLS is located in the amino acid range- 240-357. It is so because it can be observed from the available results that localization into the nucleus takes place for deletion mutants having amino acids - 1-576, 115- 576, 240-576 , 240- 357 & 115-357. Conclusions: 115-576 - NLS is not located in amino acid sequence 1-115 240-576 - NLS is not located within the amino acid sequence - 1-239 240-357 - NLS is not located in amino acid sequence 1-240 & 358-576 115-357 - NLS is not located within 1-115 & 358-576 Localization takes place in all the above cases and the deletion mutants are always found in the nucleus and they may or may not be found in the cytoplasm. From assessment of these conclusions, it can only be inferred that NLS is located in the range 240-357. Answer B: There are some “shuttling” proteins that contain a nuclear-export signal (NES) that stimulates their export from the nucleus to the cytoplasm through nuclear pores, in addition to an NLS that results in their reuptake into the nucleus. Export of proteins from the nucleus is mediated by proteins known as Exportins which first forms a complex with Ran·GTP and then binds the NES in a cargo protein. Once it crosses the nuclear pore, the Ran GAP associated with the NPC cytoplasmic filaments stimulates conversion of Ran·GTP to Ran·GDP. The accompanying conformational change in Ran leads to dissociation of the complex. The NES-containing cargo protein is released into the cytosol, while exportin 1 and Ran·GDP are transported back into the nucleus through NPCs. In this case, NES is located in the amino acid region 115-240. It is speculated to be so because: The deletion mutant bearing amino acids 115-576 possess the NES since i.

AnswerThe nucleus is separated from the cytoplasm by a phospholip.pdf

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21 Solu.pdf

1. 2:1 Solution 2:1

21 Solu.pdf

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