Which of the following would increase urine volume1) a diuretic d.pdf

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Which of the following would increase urine volume? 1) a diuretic drinking 2) large volume of water. 3) developing diabetes 4) overexpression of aquaporins Solution All listed factors can increase urine volume Diuretic agents causes diuresis, increase urine production. Large volume of water consumed leads to more urine formation Diabetes diseases symptoms includes frequent urination. Overexpression of aquaporins proteins respoponsible for water transfer can also increse urine volume.

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Is there a similarity transformation that maps Delta ABC to Delta DBF.pdf

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Felix and Janet Green live in Swarthmore, PA. Janet\'s father, Larry, lives in Sweden. For each of the following transactions that occur in their lives, identify whether it is included in the calculate investment (I), government purchases (G), exports (X), or imports (M). Check all that apply. Solution 1. It is part of investment (I) 2. Yes, It is part of imports (M) 3. Yes, It is part of consumption (C) 4. Yes, It is part of consumption (C) 5. Yes, It is part of government spendig (G).

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Do the leaves of all plants have an abscission layer? If not, what happens to the leaves on such plants? Solution Plants exhibit the phenomenon of abscission when it no longer needs its parts like leaves, flowers and fruits. Most of the plants undergo abscission to conserve nutrients and energy. Almost all plants and trees of deciduous, coniferous and rain forests shed leaves by forming abscission layer. Abscission layer formation shed the plant part with deposition of secondary structures made up of parenchyma, suberin and lignin as a protective layer. Abscission in plants take place under 3 major conditions- 1. to conserve water and photosynthetic energy; 2. to resist the pests or as part defense mechanism and 3. to improve photosynthetic efficiency of plants. Generally, the leaves that became older are shed through abscission. This is because, the older leaves lose chlorophylls and may develop other pigments like carotenes, xanthophylls and anthocyanins abundantly. If abscission does not take place, the older leaves remaining on plant may not carry out photosynthesis efficiently as a chlorophyllous leaf does. Due to this plant may exhibit poor growth and nutrient deficiency and may even die due to inadequate nutrient synthesis..

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Discuss the outcomes and challenges surrounding shortages in various markets. Below are a few examples from NBC learn. Pick one of these to discuss: Bees Tamiflu Workers Rice Solution The report notes that the record population can be viewed as a success because it means people are living longer — average life expectancy has increased from about 48 years in the early 1950s to about 68 in the first decade of the 21st century — and more children are surviving worldwide. But not everyone has benefited from a higher quality of life. In some of the poorest countries women are having more babies, stymieing development and perpetuating poverty; in some of the wealthier countries low fertility rates and a shortage of workers are raising concerns about the sustainability of economic growth and social programs. World population by the billions 1 billion - 1804 2 billion - 1927 3 billion - 1959 4 billion - 1974 5 billion - 1987 6 billion - 1999* 7 billion - 2011 *Year when the milestone was observed by the United Nations. Source: United Nations Population Fund \"This report makes the case that with planning and the right investments in people now — to empower them to make choices that are not only good for themselves but for our global commons — our world of 7 billion can have thriving, sustainable cities, productive labor forces that can fuel economic growth, youth populations that contribute to the well-being of economies and societies, and a generation of older people who are healthy and actively engaged in the social and economic affairs of their communities,\" writes Babatunde Osotimehin, executive director of the UNFPA..

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Create a C program that implements \"The Game of Life\" cellular automata (CA) simulation. A few comments: You will want 2 two-dimensional arrays. One that holds the current state of the \"universe\" and second that is generated from the first using the Game of Life rules, which are given below. After the computations for filling the second array are complete, the contents of the second array can be copied back to the first. You can make the array as big as you want, but you should probably use 20 times 20 as a minimum. As discussed in class, the outside boundary of the \"universe\" can be a bit tricky. To keep it simple for this exercise, you can assume that the outside boundary cells are always \"dead\". You can use whatever character (or number) that you want to represent the \"living\" cells. The dead cells should probably be spaces. The rules for whether a particular cell lives or dies are determined by the 8 surrounding cells. If a particular cell is alive, then: If only 0 or 1 of the surrounding cells are alive, the cell dies. (It\'s lonely.) If 4 or more of the surrounding cells are alive, the cell dies. (Too much competition.) If 2 or 3 of the surrounds cells are alive, the cell stays alive. If a particular cell is dead, then it comes to life if exactly 3 of the surround cells are alive. You will have to pick some starting configuration for the universe. An easy thing to do is fill the array randomly, and then let things go and see what evolves. Write your program as described above and demo the operation to your TA. Then try some modifications. First, try using a different starting configurations. Look here for some ideas:. Gosper\'s Glider Gun is a famous one. Or try a simple spaceship or oscillator. Then try changing the rules determining life and death. A common variation is to include a \"birth\" when 6 neighboring cells are alive in addition to the \"3-cell\" birth. Finally, there will be a short quiz covering 2D arrays. This exercise should be very familiar. Solution #include #include #include #include #include #include #include #define ROWS 60 #define COLS 60 #define OFF 0 #define ON 1 #define BLACK 8, 0, 0, 0, 0 #define scr_width 600 #define scr_height 600 #define cell_width (scr_width / ROWS) #define cell_height (scr_height / COLS) We must hold two copies of the board so that we can analyze board and make * changes to temp when killing/creating cells so that we don\'t cause changes * to the board to affect the following cells. */ char board[ROWS][COLS]; char temp[ROWS][COLS]; SDL_Rect cells[ROWS][COLS]; /* Stores positions of each cell for blits. */ void randomize_board(void); void initialize_grid(SDL_Surface* screen); void blit_board(SDL_Surface* bcell, SDL_Surface* screen); int num_neighbours(int x, int y); void update_board(void); int clear_board(SDL_Surface* screen, Uint32 color); int clear_cell(SDL_Surface* screen, int x, int y, Uint32 color); void initialize_cells_array(void); int main(void) { SDL_Init(SDL_INIT_VIDEO).

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Copy a String in Reverse Order Write a program with a loop and indirect addressing that copies a string from source to target, reversing the character order in the process. Use the following variables: source BYTE \"This is the source string\",0 target BYTE SIZEOF source DUP(\'#\') and here is what I have got so far ;reverse INCLUDE IRVINE32.INC .DATA SOURCE byte \"this is the source string\", 0 target BYTE SIZEOF source DUP(\'#\') .code main PROC Call Clrscr mov esi,0 mov edi,SIZEOF source-TYPE source mov ecx,LENGTHOF source L1: mov al, source[edi] mov target[esi] sub edi TYPE source loop L1 exit main ENDP END main Solution program with a loop and indirect addressing that copies a string from source to target MOV SI, source ; Get source address MOV DI, (target + SIZEOF source) ; Get the ending address for target LOOP: MOV AL, [SI] ; Get a byte MOV [DI], AL ; Store a byte INC SI ; Move forward one in source DEC DI ; Move back one in target CMP AL, 0 ; Have we reached the zero termination? JNZ LOOP RET.

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Compare and contrast prokaryotic cells and eukaryotic cells. Compare and contrast passive transport and active transport. Compare and contrast animal cells and plant cells. Summarize the endosymbiosis theory including the evidence that supports it. Solution 1. Similarities between prokaryotic cells and eukaryotic cells: Differences between Prokaryotic cells and Eukaryotic cells: 2. Similarities between active and passive Transport: Differentiate between Active and Passive Transport:Prokaryotic CellsEukaryotic cellsCells don\'t have nucleus.Cells have nucleus.They don\'t have organelles.They have organelles.They all are unicellular.They may be unicellular or multicellular..

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Case Study 1 Hotel worker Danny Ruiz was living with his wife and four children in a cramped New York apartment when he saw a television ad promising the family a way out. “Why rent when you can own your own home?” Penn- sylvania builder Gene Percudani asked. The company even offered to pay his rent for a year, while he saved for a down payment. So the Ruiz family fled the cityhe Pocono Mountains, where they bought a three- bedroom Cape Cod home for $171,000. However, when they tried to refinance less than two years later, the home was valued at just $125,000. “I just about flipped,” said Mr. Ruiz. Later Mrs. Ruiz remarked about her husband, “He went nuts.” Percudani, a 51-year-old native of Queens, New York, built a thriving homebuilding business in this market, running folksy television ads offering New Yorkers new homes in Pennsylvania. If they joined Percudani’s program, called “Why Rent,” homeowners would find financing through another of his companies, Chapel Creek Mort- gage, which brokered loans from J. P. Morgan Chase and the company’s Chase Manhattan Mortgage unit. For years, the “Why Rent” program appealed to workers with modest salaries, such as Eberht Rios, a truck driver for UPS. Rios bought a home in the Poco- nos for $140,000. This year, when he tried to refinance, he was told the home was valued at only $100,000. One local appraiser, Dominick Stranieri, signed off on most of the “Why Rent” deals that state officials now say were overpriced, including the Rios and Ruiz homes. Percudani’s firm picked Stranieri as his appraiser because of his quick work and low fee of $250, instead of the typical $300 to $400. In exchange for a steady stream of work, Mr. Stranieri accepted without question valuations from Percudani’s company. Other common methods of creating revenues include investors and others buying distressed properties and then, using inflated appraisals, selling them for a big profit. In order to secure the efforts of a “dirty appraiser,” those involved with the fraud would pay up to $1,500 under the table on top of the appraiser’s stan- dard fee of $400. Another unique twist to the plot is that few of the people involved in making mortgage loans have a long- term interest in them. Traditionally, bankers made loans directly and held them, giving the lenders a strong incentive to find fair appraisals to protect their interest. Today, however, many appraisers are picked by independent mortgage brokers, who are paid per transaction and have little stake in the long-term health of the loans. Many lenders have also lost a long-term interest in their loans, because they sell them off to investors. Appraisers increasingly fear that if they don’t go along with higher valuations sought by brokers, their business will dry up. Do you think a county appraiser would do a lot bet- ter than a private practitioner? Joel Marcus, a New York–based attorney recently had his property valued at $2.2 million by a county appraiser, up fro.

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Assignment Submission Assignment, you submit answers by question by question parts. The number of submissions remaining for each question part only changes if you submit or change the answer. Assignment Scoring Assignment submission is used for your score. The complex number z = -1 + i in polar form is z =. The complex number z = 2(cos(pi/6) + i sin(pi/6)) in rectangular form is z = ___________. The complex number graphed below can be expressed in rectangular form as z = or in polar form as z = _________. Solution a)given z =-1+ 1i r =[(-1)2+12] r=2 =tan-1(1/-1) =-tan-1(1/1) =-(/4) =(3/4) z=2 [cos(3/4) +i sin(3/4)] ==================================== b)z=2 [cos(/6) +i sin(/6)] z=2 [(3 /2) +i(1/2)] z=3 +i1 z=3 +i c) z =(1+ 1i) z=2 [cos(/4) +i sin(/4)].

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A 7 year old female with a history of previous UTI’s suddenly develops painful, dark-orange colored urine, with fever and back pain. Her urinalysis reveals dipstick tests that are positive for protein (2+), blood (1+), leukocyte esterase (4+), and nitrate (4+). The microscopic examination yields 20-25 WBC/HPF, 5-10 RBC/HPF, 5-10 WBC Casts/LPF, and numerous bacteria. The patient has been taking Tylenol and vitamin C supplements daily for the past 24 hours, since the condition began. Based on the results 1) what diagnosis/condition would you surmise at this point? 2) what dipstick/microscopic test support this? 3) is the patient’s history significant? 4) can the medication or vitamin supplement have any impact on the test results? Solution Women often experience urinary tract infections that are often marked by pain in the abdomen or lower back. The urine is dark-orange colored due to blood mixing with dark yellow urine. The infection may be due to bacteria E. coli, (MayoClinic). Leukocyte esterase test dipstick that indicates; the presence of white blood cells either as whole cells or as lysed cells; supports this assumption and the microscopic examination yielding 20-25 WBC/HPF also backs this assumption. Yes with all the medical history and the details of the tests done; one can proceed with the treatment. Yes according to researches; Vitamin C spiking of urine has shown to demonstrate false-negative results at various concentrations, (Ko DH, 2015) in case of Dipstick Test Results..

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5. Toxco, Inc. emits a noxious gaseous substance as a waste product of its operations. As part of its manufacturing process, it also generates liquid solvent wastes such as toluene. Analyze which if any, of these materials are regulated under RCRA. If there is any additional information that would be needed to answer this question, identify what that is. (15 points) Solution Following the discovery of an illegal meth lab by law enforcement, officers collect samples of the chemicals used in that lab as evidence for prosecution. After all forensic samples are gathered, a Hazardous Materials (Hazmat) Team is summoned to clean up the remaining chemicals and other by-products or wastes. Estimates indicate that for each pound of meth produced, between five and six pounds of highly toxic waste is generated.1 The by-products or wastes removed by the Hazmat Teams are generally treated as RCRA hazardous waste without further analysis, and are sent to a RCRA permitted site for treatment and disposal. State and Federal officials also have found that during the operation of the meth lab, the residue from meth production may enter porous building materials such as wallboard, carpets and furnishings.2 In addition to the potential contamination of the building and its contents, there may be further contamination from the improper disposal of these by-products or wastes by the producers of the meth, which is typically poured into the indoor plumbing drains that flow either into the city sewer or individual sewage treatment systems (e.g., septic systems), or modified to allow drainage directly onto the soil near the meth lab. The other disposal method typically seen at functioning meth labs is a burn or burial pit. The primary environmental concern from drainage of these materials into the soil or from disposal into a burn or burial pit is contamination of ground water by the chemicals, including volatile organic chemicals used in the production of meth, as well as vapor intrusion which involves soil vapor gas from ground water plumes infiltrating, and accumulating in, nearby buildings (residences and nonresidential sites) at dangerous levels.3 In most States, the property owner is responsible for decontamination of the property. In some States, the health department or another agency may evaluate the property for risks from long-term exposure to residual chemicals, including the potential environmental impacts of chemical spills into the environment or improper waste disposal. V. How Does RCRA Regulate Hazardous Waste? The determination of whether a particular waste is a Subtitle C RCRA hazardous waste is a multi-step process that depends on a number of factors, including the chemical character and composition of the waste. The regulations for making the hazardous waste determination are found in 40 CFR part 261.4 The first step is to determine whether the waste in question is a solid waste as defined under RCRA. The RCRA regulations require this determinatio.

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21. Two fair coins are tossed. Let E be the event not more than one head and F the event at least one of each face. Define the sample space S = {HH, HT, TH, TT) and assign probability ¼ to each simple event. a. Determine the P(E) b. Find the P(F) c. Find the P(E intersection F) d. Are E and F independent or dependent? 23. Three fair coins are tossed. Let E be the event not more than one head and F the event at least one of each face. Define the sample space S = {HHH, HHT, ..., TTT} and assign probability 1/8 to each simple event. a. Determine the P(E) b. Find the P(F) c. Find the P(E intersection F) d. Are E and F independent or dependent? Solution 21. a) E be the Event = { Not more than one Head } = {(H,T) (T,H) (T,T) } b) F be the Event = { atleast one of each face } = { (H,T) (T,H)} = Sample Space = { (H,T) (T,H) (T,T) (H,H) } c. P( E n F) = {(H,T) (T,H)} d. Dependent Note : We are suppose to answer only 1 question in time.

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2.84 Move your business here. City officials use a variety of tactics to encourage businesses to open offices in their city. One characteristic of a city that is viewed as desirable is open public space within the city limits. The New York City Open Accessible Space Information System Cooperative (OASIS) is an organization of public- and private-sector representatives that has developed an information system designed to enhance the stewardship of open space. Below are data from the OASIS Website for 12 large U.S. cities. The variables are population in thousands and total park or open space within city limits in acres. City Baltimore. Population= 651. Open space=5091 Boston. . Population= 589. Open space=4865 Chicago . Population= 2896. Open space=11645 Long beach. Population= 462. Open space=2887 Los Angeles. Population= 3695. Open space=29801 Miami. . Population= 362. Open space=1329 Minneapolis. . Population= 383. Open space=5694 New York. . Population= 8008. Open space=49854 Oakland. . Population= 399. Open space=3712 Philadelphia. . Population= 1518. Open space=10685 San Francisco. . Population= 777. Open space=5916 Washington, D.C.. Population= 572. Open space=7504 (a) Make a scatterplot of the data using population as the explanatory variable and open space as the response variable. (b) Is it reasonable to fit a straight line to these data? Explain your answer. (c) Find the least-squares regression line. Report the equation of the line and draw the line on your scatterplot. (d) What proportion of the variation in open space is explained by population? Solution.

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1. Receptors are where transduction occurs in a sensory system. For the visual system, the receptors are called photoreceptors. There are two types of photoreceptors. What are they called? 2. What is transduction? Use your own words (not the Chapter Outline’s or the textbook’s) to give a definition as you understand it. 6. Receptors are where transduction occurs in a sensory system. For the visual system, the receptors are called photoreceptors. There are two types of photoreceptors. What are they called? 7. What is transduction? Use your own words to give a definition as you understand it Solution 1.The two types of photo receptors seen are Rods and Cones. 2. Transduction in lay man language refers to transfer of foreign DNA to some specific cell by means of vector. Generally here the vector used is Virus. 6. Please refer to my answer in question 1. it sates that rodes and cones are two photoreceptors. 7. same as the answer in question 2..

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-acute appendicitis (inflammation of the vermiform appendix) requires immediate surgical removal of the appendix. in the event that the appendix bursts, its contents are released into the abdominal cavity, which can lead to widespread infection, sepsis, and death. discuss the events that might lead to an enlarged and inflamed vermiform appendix. -following the 2009 H1N1 pandemic, annual influenza vaccines began to include strains for H1N1 influenca. Discuss the purpose of vaccines and state how vaccines may protect against disease outbreak. Solution appendicitis is obstruction of the lumen of the appendix followed by infection. In 60% of patients, obstruction is caused by hyperplasia of the submucosal follicles. This form of obstruction is mostly observed in children and is known as catarrhal appendicitis. A fecalith or fecal stasis causes luminal obstruction 35% of the time and is usually observed in adults. Obstruction may also be caused by foreign bodies (4%) and tumors (1%). Following obstruction, an increase in mucous production occurs, leading to increased intraluminal pressure. With increased pressure and stasis from obstruction, bacterial overgrowth ensues. The mucus then turns into pus that causes a further increase in luminal pressure. This leads to distention of the appendix and visceral pain, which is typically located in the epigastric or periumbilical region. Purpose of the vaccine: The purpose of vaccination is to produce immunity against particular disease . Many infectious diseases, previously responsible for millions of deaths, are being controlled through vaccination. Some, such as smallpox, polio have been eradicated through vaccination. How vaccines protect against diseases? Vaccination comes under passive immunisation , where already existed antibodies are extracted from other organisms which are infected with the disease. These antibodies injected into the persons body to give immunity against the same disease in future..

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How terrorism may be prevented by new technologies? Solution Terrorism may be prevented by implementing new technologies ,but how they can be prevented ..? To prevent terrorism we should be improved in specific areas of technologies 1)Telecommunication:it is one of the main area to improve our technologies because lot dealing are done threw phone call if we analyse the calls then we can easily prevent them .do we need good and efficient analysis of data in telecommunication 2)gps: we not up to mark to get the Geo location at more accurately we need to keep an eye from the sky to catch each and every piece of the earth which makes easier to find their dens 3)weapons techniques:we need to improve our weapons which point out with more accuracy.

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Which of the following are mechanisms of carrier mediated transport? singular transport Locomotive transport Membrane-bound transport Passive transport Active transport Which of the following are properties of simple diffusion? Solute export is highly selective Rate of solute entry is dependent on the concentration gradient Solute entry is highly specific Rate of solute entry is saturable Rate of solute entry is dependent on the cell\'s surface area I am a long and thin structure made of protein that mediates swimming motility. What cell structure am l? Capsule Pilus Fimbriae Flagellum slime layer Solution 3. the mechanism which follows carrier mediated transport is membrane bound transport 4.rate of solute entry is dependent on the concentration gradient 5. flagellum.

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Which of the following lists all the types of molecules that could be found in the plasma membrane of animal ceils? Phospholipids, proteins, glycoproteins, glycolipids, cholesterol, sphingolipids. triacylglycerides, Phospholipids, proteins, glycoproteins, glycolipids, cholesterol, sphingolipids Phospholipids, proteins and cholesterol Phospholipids, proteins, glycolipids, cholesterol, triacylglycerides Phospholipids, proteins, glycolipids, cholesterol, sphingolipids Which of the following is true about the phospholipids one would observe in the membranes of an eukaryotic organism? All the membranes in the organism including the membranes of organelles of cells would contain the same ratio of different phospholipids. The type of phospholipids found in the membranes of a cell would vary by the cell type but would be the same within all the organelles of the cell. The type of phospholipids found in the membrane of a cell would vary in the different cell types, in the membranes of the different organelles and in the two halves of each membrane in the different parts of the cell. The type of phospholipids found in the membranes of a cell would vary in the different cell types and in the organelles of the different cell types but would be the same in both halves of membrane in each part of a cell. All the membranes in the organism including the membranes of organelles would contain only one type of phospholipid. A feature common to almost all transmembrane proteins is an a-helical region of about 20 to 25 hydrophobic amino acids. a structure consisting almost exclusively of beta-sheets. an amino acid sequence rich in acidic residues. a phosphorylated exterior domain. Active transport differs from facilitated diffusion in that active transport involves a conformational change in the transport molecule. active transport involves the transport of molecules up or against their concentration gradient. ions are not transported via active transport. active transport requires a protein component, whereas facilitated diffusion occurs by simple diffusion through the plasma membrane. Solution 5. ANS: A. phospholipid, Proteins, glycol proteins, glycol lipids, cholesterol, spingolipids and tri-acyl glycerol. Explanation: By using these lipids and proteins (phospholipid, Proteins, glycol proteins, glycol lipids, cholesterol, spingolipids and tri-acyl glycerol) animal cell membrane serves as a barrier to determine whether what things can enter or leave in organelle. 6. ANS: D. The type of the phospholipids fond in the membrane of the cell would vary in the different cell types but would be the same in both halves of membrane in each part of a cell. 7. ANS: A. An alfa helix region of about 20-25 hydrophobic amino acids. Explanation: Most Trans membrane proteins are usually have one or more Alfa helices in membrane bilayer; an alfa region contains 20 to 25 hydrophobic amino acids. 8. ANS: B. Active transport involves the transport of molecules up or against their co.

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