Which of the following lists all the types of molecules that could be.pdf

•

0 likes•2 views

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.

Education

an alfa region contains 20 to 25 hydrophobic amino acids.
8. ANS: B. Active transport involves the transport of molecules up or against their concentration
gradient.
Explanation: Cell use active transport to get needed molecules regard less of concentration
gradient to maintain homeostasis.

Similar to Which of the following lists all the types of molecules that could be.pdf

Structure of cell membrane and Transport

Mohit Adhikary

HMP01 02Membrane structure of excitable cell (1).pdf

kennedymaweu032

Membrane slideshare 2014

jrfisher78

The Endomembrane System

Nestor Lasala Jr.

THE CELL PART 2.pptx

SbonakalisoGumede

CELL ORGANELLES

DEV YADAV

Molecular Cell Biology the endoplasmic reticulum

HammadBhatti20

cell membrane supriya

Supriya hiremath

The endomembrane system

Bahauddin Zakariya University lahore

Lysosomes mbb

AshishNain

easylearningwithned.blogspot.com-FLUID MOSAIC MODEL.pdf

Home

MEMBRANE STRUCTURE

Dobbs Ferry High School

Clase 5_Biomembranas y Bioseñalización (1).pdf

KarolLuciaChajinNarv1

Structure of plasma membrane

HARINATHA REDDY ASWARTHA

Fluid- mosaic- model.pptx

beherasudarshan187

2 + 3. cell biology introduction + fluid mosaic model

Azhen Qadir

Structure of plasma membrane

shilpa nair

Plasma membrne

Dilip Pandya

Cell membrane

Akanksha Golchha

2. Absorption & Secretion Of Materials

rossbiology

Similar to Which of the following lists all the types of molecules that could be.pdf (20)

Structure of cell membrane and Transport

HMP01 02Membrane structure of excitable cell (1).pdf

Membrane slideshare 2014

The Endomembrane System

THE CELL PART 2.pptx

CELL ORGANELLES

Molecular Cell Biology the endoplasmic reticulum

cell membrane supriya

The endomembrane system

Lysosomes mbb

easylearningwithned.blogspot.com-FLUID MOSAIC MODEL.pdf

MEMBRANE STRUCTURE

Clase 5_Biomembranas y Bioseñalización (1).pdf

Structure of plasma membrane

Fluid- mosaic- model.pptx

2 + 3. cell biology introduction + fluid mosaic model

Structure of plasma membrane

Plasma membrne

Cell membrane

2. Absorption & Secretion Of Materials

More from arihantsherwani

The age of the Earth is estimated from the amount of uranium isotope 238 remaining. Radioisotopes like U^238 slowly decay over time turning into either another isotope or another element when they release radiation. Estimates are that only half of the U^238 that was originally present in the Earth still exists. Since the half life (time it takes a radioisotope to lose 1/2 its material) is 4 5 times 10^9 that puts the age of the Earth at 4 5 billion years U^238 decays into Th^234, which decays into Pa^234, which decays into U^234. The earliest fossil evidence of bacteria is 3 5 billion years old. Vascular plants began to appear 400 million years ago. Modern humans are estimated to have appeared about 200,000 years ago. At which of these times would the most original U^238 remain? The time bacteria first became fossilized The time vascular plants appeared The time modem humans appeared At which of these times would the least original U^238 remain? The time bacteria first became fossilized The time vascular plants appeared The time modem humans appeared Fossils found in rock suggest that: Compare these time spans by changing the scale from YEARS to SECOND: let 1 year = 1 sec. Thus, in this scale the average biology student is 20 seconds old. At this scale, how many days ago did humans appear? 1.7 hours 1 day 1.7 days 2 days 2.7 days How many years ago did the dinosaurs vanish? 2 years 2.1 years 2.2 days 2.3 days 2.4 days How many years ago did flowering plants begin to dominate? 1.5 years 2 years 2.5 years 3 years 3.5 years How many years ago did plants start to live on land? 10 years 11 years 12 years 13 years 14 years How many years ago did cells first fossilize? 91 years 111 years 121 years 131 years 141 years Solution a) Most U238 existed when the bacteria first bacame fossilised. The earliest fossil evidence of bacteria is 3.5 billion years ago b) Least U238 remains at time when first modern human appeared 200,000 years ago 2a) 180,000 years = 180,000 sec = 2.08 days or 2 days. Ans iv b) 2.06 years c)2.5 years d) 12.6 years e)111 years.

The age of the Earth is estimated from the amount of uranium isotope .pdf

arihantsherwani

Suppose Michael creates an RSA cryptosystem with a very large mod ulus n for which the factorization of n cannot be found in a reasonable amount of time. Suppose that Nikita sends messages to Michael by representing each alphabetic character as an integer between 0 and 26 (A corresponds to 1, B to 2, etc., and a space to 0), then encrypts each number separately using Michael\'s RSA cryptosystem. Is this method secure? Explain your answer. Solution This method is not secure, it is just equal to a \"Ceaser Cipher\", that is a permutation of letters of alphabet, which is well-known to be easily broken using a frequency analysis..

Suppose Michael creates an RSA cryptosystem with a very large modulu.pdf

arihantsherwani

Match the terms to the definitions provided. Question 2 options: dehydrogenations anaerobic respiration fermentation digestion oxidative phosphorylation heterotrophs aerobic respiration ATP synthase autotrophs substrate-level phosphorylation glycolysis High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is oxygen An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP to ATP ATP synthetase makes ATP with energy from a proton gradient generated by an electron transport chain. Organisms that live on the organic compounds produced by other organisms. The use of inorganic molecules as final electron acceptors for an electron transport chain. A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels phosphorylation. Reactions in which electrons lost are accompanied by protons. ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing intermediate. The enzymatic breakdown of large molecules into smaller ones. The electrons generated by glycolysis are donated to organic molecules and NAD+ is recycled Organisms that harvest energy from the sun and convert it to chemical energy by photosynthesis. dehydrogenations anaerobic respiration fermentation digestion oxidative phosphorylation heterotrophs aerobic respiration ATP synthase autotrophs substrate-level phosphorylation glycolysis1. High-energy electrons from chemical bonds are transferred to a final electron acceptor, which is oxygen2. An enzyme that couples the reentry of protons into the matrix with the phosphorylation of ADP to ATP3. ATP synthetase makes ATP with energy from a proton gradient generated by an electron transport chain.4. Organisms that live on the organic compounds produced by other organisms.5. The use of inorganic molecules as final electron acceptors for an electron transport chain.6. A series of 10 reactions used to catabolize glucose and produce ATP by substrate levels phosphorylation.7. Reactions in which electrons lost are accompanied by protons.8. ATP is formed by transfer of a phosphate group directly to ADP from a phosphate-bearing intermediate.9. The enzymatic breakdown of large molecules into smaller ones.10. The electrons generated by glycolysis are donated to organic molecules and NAD+ is recycled11. Organisms that harvest energy from the sun and convert it to chemical energy by photosynthesis. Solution Answer : Dehydrogenations : Reactions in which electrons lost are accompanied by protons. Anaerobic respiration : The use of inorganic molecules as final electron acceptors for an electron transport chain. Fermentation : The electrons generated by glycolysis are donated to organic molecules and NAD+ is recycled Digestion : The enzymatic breakdown of large molecules into smaller ones. oxidative phosphorylation : ATP synthetase makes ATP with energy from a proton gradient generated by an electron transport chain. Heterotrophs : Organi.

Match the terms to the definitions provided.Question 2 optionsd.pdf

arihantsherwani

Simple linear regression and multiple regression are both based on the use of quantitative variables to predict a single dependent variable. It is the number of independent variables that changes. Describe a regression relationship using multiple independent variables. Solution Consider Y: weight Let X1: gender X2: age X3: height X4: nature of job X5: heredity factors This is one example where multiple regression is used..

Simple linear regression and multiple regression are both based on t.pdf

arihantsherwani

Sand, Mell, and Rand are partners who share incomes and losses in a 1:4:5 ratio. After lengthy disagreements among the partners and several unprofitable periods, the partners decided to liquidate the partnership. Before the liquidation, the partnership balance sheet showed the following: Cash $10,000 Total “other assets,” $106,000 Total liabilities, $88,000 Sand, Capital, $1,200 Mell, Capital, $11,700 Rand, Capital, $15,100 The “other assets” were sold for $ 85,000. Proceeds from the sale of other assets were used to payoff existing liabilities. Determine the following: 1)The gain (or loss) realized on the sale of the assets and recording of liabilities’ payment. 2)The balances in the partners’ capital accounts after the distribution of this gain or loss to the capital accounts. 3)Assume that if any capital deficits exist, they are not made up and the deficient partner pays down his or her deficit to zero. How much cash will each of the partners receive in the final liquidation? Solution Balance Sheet Assets Amount Liabilities Amount Cash $ 10,000.00 Total liabilities $ 88,000.00 Other Assets $ 106,000.00 Partner\'s Capital A/c Sand $ 1,200.00 Mell $ 11,700.00 RAND $ 15,100.00 Total $ 116,000.00 Total $ 116,000.00 Journal Entries Particular Amount(DR) Amount(CR) 1) Cash A/c $ 85,000.00 Loss on realization $ 21,000.00 To Other Assets $ 106,000.00 (Being amount realized on sale of assets) Sand=($21000*1/10) $ 2,100.00 Mell=($21000*4/10) $ 8,400.00 Rand=($21000*5/10) $ 10,500.00 To Loss on realization $ 21,000.00 (Being loss on realization distribute to partner\'s in their profit sharing ratio) Total Liabilities $ 88,000.00 To Cash $ 88,000.00 (Being amount paid to total liabilities) 2) Sand\'s Capital A/c Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $ 2,100.00 By Balance B/d $ 1,200.00 By Cash A/c $ 900.00 Total $ 2,100.00 Total $ 2,100.00 Mell\'s Capital A/c Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $ 8,400.00 By Balance B/d $ 11,700.00 To Cash $ 3,300.00 Total $ 11,700.00 Total $ 11,700.00 Rand\'s Capital A/c Particular Amount(Dr) Particular Amount(Cr) To Loss on realization $ 10,500.00 By Balance B/d $ 15,100.00 To Cash $ 4,600.00 Total $ 15,100.00 Total $ 15,100.00 3) Cash A/c Particular Amount(Dr) Particular Amount(Cr) To Balance b/d $ 10,000.00 By Total Liabilities $ 88,000.00 To Other assets $ 85,000.00 By Mell\'s Capital A/c $ 3,300.00 To Sand\'s Capital A/c $ 900.00 By Rand\'s Capital A/c $ 4,600.00 Total $ 95,900.00 Total $ 95,900.00.

Sand, Mell, and Rand are partners who share incomes and losses in a .pdf

arihantsherwani

PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE. THANK YOU! Why do all (-)RNA virus have an RNA polymerase in its virion? How does vesicular stomatitis virus regulate its mRNA transcription? How does vesicular stomatitis virus regulate Protein synthesis? State numerous factors for Ebola virus being an emerging deadly pathogen for us. Why can’t influenza virus replicate in a cell where the nucleus has been removed? PLEASE ANSWER ALL THE QUESTIONES ABOVE AS DETAILED AS POSSIBLE. THANK YOU! Solution 1) Negative-sense RNA has a nucleotide sequence complementary to the mRNA that it encodes. Like DNA, this RNA cannot be translated into protein directly. Instead, it must first be transcribed into a positive-sense RNA that acts as an mRNA. In order to get transcribed into a positive-sense RNA these viruses need a RNA Polymerase enzyme. As soon as the virion enters the host cell the virus uses its own RNA replicase, known as RNA dependent RNA Polymerase to form positive RNA template strands through complementary base pairing. This positive sense RNA strand acts as mRNA and can be translated into structural capsomere proteins and viral RNA polyemrase enzymes for further synthesis of virions. 2) The RNA-dependent RNA polymerase of vesicular stomatitis virus, a nonsegmented negative- strand RNA virus, directs two discrete RNA synthetic processes, transcription and replication. The two short extragenic regions at the genomic termini, the 3 leader (Le) and the complement of the 5 trailer (TrC), contain essential signals for these processes. The viral genome comprises a 47-nt leader region (Le), five transcriptional units for the N, P, M (matrix), G (glycoprotein), and L mRNAs, and a 59-nucleotide trailer region (Tr), arranged in the order 3 Le-N-P-M-G-L-Tr 5. The 3 Le is thought to act as a promoter for transcription of the viral mRNAs. The promoter for mRNA transcription was shown to include specific sequence elements within Le at positions 19 to 29 and 34 to 46, a separate element at nucleotides 47 to 50, the nontranscribed leader-N gene junction. Transcription process 3) Translation in Vesicular Stomatitis Virus 3) Ebola, commonly known as Ebola hemorrhagic fever, is a rare and deadly disease caused by infection with one of the Ebola virus strains. Ebola can cause disease in humans and nonhuman primates (monkeys, gorillas, and chimpanzees). Ebola viruses belong to a family of viruses called Filoviridae, and there are five known species of Ebola virus 5) Influenza A virus belongs to the family of Orthomyxoviridae. Orthomyxoviridae viruses are one of two RNA viruses that replicate in the nucleus (the other being retroviridae). The reason is that the machinery of orthomyxo viruses cannot make their own mRNAs. The influenza virus has a negative sense RNA genome. In the host nucleus, the virus does primary transcription to produce necessary proteins for replication..

PLEASE ANSWER ALL THE QUESTIONES BELOW AS DETAILED AS POSSIBLE. THAN.pdf

arihantsherwani

Livingstone is a marble freak and loves to paly at school. During the first break, he doubled the number of marbles he had. During the second break, he lost 4 marbles. During the third break, he increased his stock of marbles by one-third. During lunch, however, he lost half of his marbles. After school, livingston played again and tripled what he had. He went home with 72 marbles. How much did he start with? work backword to answer this question. Solution consider he had x marblles During the first break, he doubled the number of marbles he had. = 2x During the second break, he lost 4 marbles =2x-4 During the third break, he increased his stock of marbles by one-third =2x-4 +1/3 (2x-4) During lunch, however, he lost half of his marbles. = (2x-4 +1/3 (2x-4))/2 After school, livingston played again and tripled what he had = 3(2x-4 +1/3 (2x-4))/2 He went home with 72 marbles 3(2x-4 +1/3 (2x-4))/2 = 72 (3/2)(4/3)(2x-4) = 72 4x-8 =72 4x = 80 x=20 he start with 20 marbles.

Livingstone is a marble freak and loves to paly at school. During th.pdf

arihantsherwani

Invertebrates colonized terrestrial environments independently of vertebrates. However, they faced similar challenges adapting to land. Compare the adaptations of flatworms, mollusks and arthropods in regards to: i) water conservation, 2) movement/body support and 3) reproduction that allowed them to successfully colonize terrestrial environments. Solution Free-living flatworms are carnivores and scavengers that eat organic matter,which they acquire through their pharynx.Tapeworms attach to their host’s intestine and absorb food through the body wall.Flatworms move by contracting muscles against a hydrostatic skeleton.Flatworms reproduce asexually by pinching in half;they can also reproduce sexually inside a host by producing eggs that are dispersed.The flat body shape of a flatworm maximizes the surface area in contact with the environment. Mollusks have a mantle that secretes a shell in most species,a muscular foot that provides movement,a visceral mass containing the internal organs,and a radula.Mollusks feed by filtering organic particles out of the water;by scraping algae off rocks;or by hunting prey.Mollusks move by contracting muscles against a hydrostatic skeleton.Many mollusks protect themselves with a hard shells,others can evade predators by changing color and darting away,often in a cloud of ink. Arthropods use their jointed appendages for movement,defense,prey capture,sensing the environment,ornaments,weapons,and copulation.Arthropods feed on almost every imaginable inorganic substance,either by chewing or sucking.On land,arthropods exchange gases via tubules called tracheae or via book lungs;arthropods that live in water respire with gills.Arthropods have specialized organs that remove nitrogenous wastes,which are deposited into the end of the digestive tract and ejected though the anus.Arthropods have specialized sense organs that are sensitive to light,sound,scents,chemicals, potential mates,food proximity,touch,air currents,and body position.Arthropods move by contracting muscles attached internally to the exoskeleton.Although there are asexually reproducing species, most arthropods reproduce sexually,using internal fertilization.The exoskeleton offers some protection against predators, but many arthropods defend themselves by fleeing,biting,stinging,pinching,making noises,or giving off foul odors.Camouflage is another defense,as are wing spots that may startle predators.Aquatic arthropods respire with gills.Terrestrial forms rely on diffusion through tiny tubes called trachea.Trachea are cuticle-lined air ducts that branch throughout the body,and open in tiny holes called spiracles,located along the abdomen.Insects can open and close these spiracles,to conserve water that would otherwise be lost to evaporation from the open tubes.Their reliance on diffusion for respiration is one of the reasons that insects are small.Being small,an arthropod has a larger relative surface area than a human, so it loses more water.The exoskele.

Invertebrates colonized terrestrial environments independently of ve.pdf

arihantsherwani

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

arihantsherwani

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).

Felix and Janet Green live in Swarthmore, PA. Janets father, Larry,.pdf

arihantsherwani

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..

Do the leaves of all plants have an abscission layer If not, what h.pdf

arihantsherwani

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..

Discuss the outcomes and challenges surrounding shortages in various.pdf

arihantsherwani

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).

Create a C program that implements The Game of Life cellular auto.pdf

arihantsherwani

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.

Copy a String in Reverse Order Write a program with a loop and ind.pdf

arihantsherwani

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..

Compare and contrast prokaryotic cells and eukaryotic cells. Compare.pdf

arihantsherwani

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 proper- ties 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.

Case Study 1Hotel worker Danny Ruiz was living with his wife and f.pdf

arihantsherwani

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)].

Assignment Submission Assignment, you submit answers by question by .pdf

arihantsherwani

A linear dynamical system can be created for two masses connected by.pdf

arihantsherwani

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..

A 7 year old female with a history of previous UTI’s suddenly develo.pdf

arihantsherwani

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.

5. Toxco, Inc. emits a noxious gaseous substance as a waste product o.pdf

arihantsherwani

More from arihantsherwani (20)