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3905.03 gm Solution 3905.03 gm.

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The possible causative agent is Corynebacterium diptheriae Sore throat, fever, shaking and chills, difficulty in breathing are all symptoms typically assocaited with C. diptheriae. The bacterium is Gram positive bacillus, non capsulated, non motile, toxin producing. They produce exotoxin called diptheria toxin, which is encoded by a prophage. The bacteria adheres to mucosal epithelium and infect them. The toxtosin released by endosomes stimulate localized inflammation followed by tissue destruction. Necrosis results from tissue destruction. The toxin belongs to AB toxin class. The B fragment binds to host cell and the A fragment toxin is innserted into the membrane and transported. It is proteolytically activated inside the cell. The toxin binds to a histidine residue of elongation factor 2 or eEF2. The active A fragment inhibits translocation of growing peptide chain on the ribosome. Thus, it inhibits protein synthesis. The toxin transfers NAD to dipthamide (elongation factor 2) to inactivate the elongation factor. Local tissue destruction makes the bacteria gain access into the circulation, anters the lymphatic and blood stream, to spread to other parts of the body. Systemic spreading causes diptheria toxin act on vital organs like kidney, myocardium, and nervous system. Innate immune responses against the bacterium includes Macrophages as the first line of defense. They use pattern specific receptors to attach and engulf bacteria. Neutrophils also mediate the same response. Macrophages further release cytokines to mediate inflammation. Solution The possible causative agent is Corynebacterium diptheriae Sore throat, fever, shaking and chills, difficulty in breathing are all symptoms typically assocaited with C. diptheriae. The bacterium is Gram positive bacillus, non capsulated, non motile, toxin producing. They produce exotoxin called diptheria toxin, which is encoded by a prophage. The bacteria adheres to mucosal epithelium and infect them. The toxtosin released by endosomes stimulate localized inflammation followed by tissue destruction. Necrosis results from tissue destruction. The toxin belongs to AB toxin class. The B fragment binds to host cell and the A fragment toxin is innserted into the membrane and transported. It is proteolytically activated inside the cell. The toxin binds to a histidine residue of elongation factor 2 or eEF2. The active A fragment inhibits translocation of growing peptide chain on the ribosome. Thus, it inhibits protein synthesis. The toxin transfers NAD to dipthamide (elongation factor 2) to inactivate the elongation factor. Local tissue destruction makes the bacteria gain access into the circulation, anters the lymphatic and blood stream, to spread to other parts of the body. Systemic spreading causes diptheria toxin act on vital organs like kidney, myocardium, and nervous system. Innate immune responses against the bacterium includes Macrophages as the first line of defense. They use pattern specific recep.

The possible causative agent is Corynebacterium diptheriaeSore thr.pdf

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The answer is E) 1,2, and 3. The solubility of a gas in solvents depends on their intermolecular interactions, which is affected by the nature of gas and the solvent as well as the temperature. So all three factors are relevant. Solution The answer is E) 1,2, and 3. The solubility of a gas in solvents depends on their intermolecular interactions, which is affected by the nature of gas and the solvent as well as the temperature. So all three factors are relevant..

The answer is E) 1,2, and 3.The solubility of a gas in solvents de.pdf

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RainfallTest.java import java.util.Arrays; import java.util.Sc.pdf

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by taking p1,p2,p3 as points in cordinate system.. displavement can .pdf

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import java.util.*; public class DecimalToBinary { public static int[] printBinaryFormat(int number){ int binary[] = new int[7]; int index = 0; while(number > 0){ binary[index++] = number%2; number = number/2; } return binary; } public static void main(String a[]){ DecimalToBinary dtb = new DecimalToBinary(); int a1[]=new int[7]; int b1[]=new int[7]; Scanner sc=new Scanner(System.in); System.out.println(\"Enter First Number\"); int m=sc.nextInt(); System.out.println(\"Enter Second Number\"); int n=sc.nextInt(); a1=printBinaryFormat(m); b1=printBinaryFormat(n); int t=sum(a1,b1); System.out.println(); if(t>0) System.out.println(\"The Sum is \"+t); else System.out.println(\"Overflow\"); } static int sum(int x[],int y[]) { int carry=0; int s1,s2,a,i; int r[]=new int[8]; i=0; int c=0; while(i<7) { a=carry+x[i]+y[i]; if(a==3) { r[c]=1; carry=1; } else if(a==2) { r[c]=0; carry=1; } else if(a==1) { r[c]=1; carry=0; } else { r[c]=0; carry=0; } i++; c++; } r[c]=carry; if(carry==1) return -1; else { int e=0; System.out.println(\"Result Number in Binary\"); for(i=6;i>=0;i--) System.out.print(r[i]); for(i=6;i>=0;i--) e+=r[i]*Math.pow(2,i); return e; } } } Solution import java.util.*; public class DecimalToBinary { public static int[] printBinaryFormat(int number){ int binary[] = new int[7]; int index = 0; while(number > 0){ binary[index++] = number%2; number = number/2; } return binary; } public static void main(String a[]){ DecimalToBinary dtb = new DecimalToBinary(); int a1[]=new int[7]; int b1[]=new int[7]; Scanner sc=new Scanner(System.in); System.out.println(\"Enter First Number\"); int m=sc.nextInt(); System.out.println(\"Enter Second Number\"); int n=sc.nextInt(); a1=printBinaryFormat(m); b1=printBinaryFormat(n); int t=sum(a1,b1); System.out.println(); if(t>0) System.out.println(\"The Sum is \"+t); else System.out.println(\"Overflow\"); } static int sum(int x[],int y[]) { int carry=0; int s1,s2,a,i; int r[]=new int[8]; i=0; int c=0; while(i<7) { a=carry+x[i]+y[i]; if(a==3) { r[c]=1; carry=1; } else if(a==2) { r[c]=0; carry=1; } else if(a==1) { r[c]=1; carry=0; } else { r[c]=0; carry=0; } i++; c++; } r[c]=carry; if(carry==1) return -1; else { int e=0; System.out.println(\"Result Number in Binary\"); for(i=6;i>=0;i--) System.out.print(r[i]); for(i=6;i>=0;i--) e+=r[i]*Math.pow(2,i); return e; } } }.

import java.util.; public class DecimalToBinary { public stat.pdf

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i did not get itSolutioni did not get it.pdf

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Hello!!!!!!! This answer will help you :) H2Se would occur in a bent formation similar to water, H2O. The asymmetry of this structure causes the molecule to be polar. The geometry of BeCl2 is linear with a symmetric charge distribution. Therefore this molecule is non polar. NF5 is a non polar. Co2 It is a non-polar molecule. But it has polar covalent bonds between its atoms. Solution Hello!!!!!!! This answer will help you :) H2Se would occur in a bent formation similar to water, H2O. The asymmetry of this structure causes the molecule to be polar. The geometry of BeCl2 is linear with a symmetric charge distribution. Therefore this molecule is non polar. NF5 is a non polar. Co2 It is a non-polar molecule. But it has polar covalent bonds between its atoms. .

Hello!!!!!!! This answer will help you ) H2Se would occur in a .pdf

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Here is the code for you: import java.util.Scanner; import java.util.Random; public class TicTacToeGame { static char[] [] board = new char[3][3]; static Scanner input=new Scanner(System.in); //Object of Stats class to maintain statistics static Stats stat = new Stats(); /** * Prints the TicTacToe board * @param arr: The board so far */ public static void printBoard(char [][] arr){ System.out.println(); for (int i=0; i<3; i++) { for (int j=0; j<3; j++) { System.out.print(arr[i][j]); if(j!=2) //Print the | for readable output System.out.print(\" \" + \"|\" + \" \"); } System.out.println(); if(i!=2) { System.out.print(\"_ _ _ \"); // Print _ for readability System.out.println();; } } } /** * Clear the TicTacToe board before starting a new game * @param arr: The board so far */ public static void clearBoard(char [][] arr){ for (int i=0; i<3; i++) { for (int j=0; j<3; j++) { arr[i][j]=\' \'; } } } /** Determines if the player with the specified token wins * * @param symbol: Specifies whether the player is X or O * @return true if player has won, false otherwise */ public static boolean isWon(char symbol) { for (int i = 0; i < 3; i++) //horizontal if (board[i][0] == symbol && board[i][1] == symbol && board[i][2] == symbol) { return true; } //TODO!!! Also check for vertical and the two diagonals for (int i = 0; i < 3; i++) //vertical if (board[0][i] == symbol && board[1][i] == symbol && board[2][i] == symbol) { return true; } //Leading diagonal if (board[0][0] == symbol && board[1][1] == symbol && board[2][2] == symbol) { return true; } //Trailing diagonal if (board[0][2] == symbol && board[1][1] == symbol && board[2][0] == symbol) { return true; } return false; } /** Determines if the cell is occupied * * @param row: Row of the cell to be checked * @param col: Column of the cell to be checked * @return true if the cell is occupied, false otherwise */ public static boolean isOccupied(int row, int col){ if (board[row][col]!=\' \') return false; else return true; } /** Determines who starts the game */ public static int whoStarts(){ //TODO: Randomly chooses between 0 and 1 and returns the choice return (int)(Math.random() + 0.5 ); } /** takes care of the human\'s move * 1. Prompt for a cell, then column * 2. Puts a symbol (X or O) on the board * 3. Prints the updated board * 4. If a human wins: prints, updates stats and returns true * 5. If not a win yet, returns false */ public static boolean humanTurn(char symbol){ //Prompt for a cell. User must enter //row and column with a space in between. System.out.print(\"\ \ Enter your move: (row column): \" ); int row = input.nextInt(); int col = input.nextInt(); //TODO!!! Mark user move in the board, print //the board and check if user has won! board[row][col] = symbol; printBoard(board); if(isWon(symbol)) return true; return false; } /** takes care of the computer\'s move * 1. Generates numbers until finds an empty cell * 2. Puts a symbol (X or O) on the board * 3. Prints the updated board * 4. If a comp .

Here is the code for youimport java.util.Scanner; import java.u.pdf

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Following are the changes mentioned in bold in order to obtain the required result and stop scrolling in the background. #include \"SDL/SDL.h\" #include //The attributes of the screen can be defined as follows const int SCN_WIDTH = 640; const int SCN_HEIGHT = 480; const int SCN_BPP = 32; //BPP defines bits per pixel SDL_Surface* Background = NULL; SDL_Surface* SpriteImage = NULL; SDL_Surface* Backbuffer = NULL; int SpriteFrame = 0; int FrameCounter = 0; const int MaxSpriteFrame = 12; const int FrameDelay = 2; int BackgroundX = 0; SDL_Surface* LoadImage(char* fileName); bool LoadFiles(); void FreeFiles(); void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y); void DrawImageFrame(SDL_Surface* image, SDL_Surface* destSurface, int x, int y, int width, int height, int frame); bool ProgramIsRunning(); int main(int argc, char* args[]) { if(SDL_Init(SDL_INIT_EVERYTHING) < 0) { printf(\"Failed to initialize SDL!\ \"); return 0; } Backbuffer = SDL_SetVideoMode(800, 600, 32, SDL_SWSURFACE); SDL_WM_SetCaption(\"Image Animation\", NULL); if(!LoadFiles()) { printf(\"Failed to load all files!\ \"); FreeFiles(); SDL_Quit(); return 0; } while(ProgramIsRunning()) { //Update\'s the sprites frame FrameCounter++; if(FrameCounter > FrameDelay) { FrameCounter = 0; SpriteFrame++; } if(SpriteFrame > MaxSpriteFrame) SpriteFrame = 0; //Background scrolling can be removed from this position //Render the scene DrawImage(Background,Backbuffer, BackgroundX, 0); DrawImage(Background,Backbuffer, BackgroundX+800, 0); DrawImageFrame(SpriteImage, Backbuffer, 350,250, 150, 120, SpriteFrame); SDL_Delay(20); SDL_Flip(Backbuffer); } FreeFiles(); SDL_Quit(); return 0; } SDL_Surface* LoadImage(char* fileName) { SDL_Surface* imageLoaded = NULL; SDL_Surface* processedImage = NULL; imageLoaded = SDL_LoadBMP(fileName); if(imageLoaded != NULL) { processedImage = SDL_DisplayFormat(imageLoaded); SDL_FreeSurface(imageLoaded); if(processedImage != NULL) { //Here we map the color key Uint32 colorKey = SDL_MapRGB(processedImage->format, 0, 0xFF, 0xFF); //Now, set all the pixels of color R 0,G 0*FF,B 0*FF to be transparent SDL_SetColorKey(processedImage, SDL_SRCCOLORKEY, colorKey); } } return processedImage; } bool LoadFiles() { Background = LoadImage(\"graphics/background.bmp\"); if(Background == NULL) return false; SpriteImage = LoadImage(\"graphics/bat.bmp\"); //The file should be preloaded and linked with the required libraries in SDL if(SpriteImage == NULL) return false; else return true; } void FreeFiles() { SDL_FreeSurface(Background); SDL_FreeSurface(SpriteImage); } void DrawImage(SDL_Surface* image, SDL_Surface* destSurface, int x, int y) //A temporary rectangle is used to hold the offsets { SDL_Rect destRect; //Giving the offsets to the rectangle destRect.x = x; destRect.y = y; //Blit the surface SDL_BlitSurface(image, NULL, destSurface, &destRect); } //Here, we need to start the main function: int main(int argc,char** args) //Now, initialize all SDL subsystems if .

Following are the changes mentioned in bold in order to obtain the r.pdf

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During meiosis, each member of a pair of genes tends to be randomly distributed into gametes (receive alleles) independently of how other chromosomes are distributed. Genes that are having their loci nearer to each other are not generally separated during chromosomal crossover and are inherited together to the offspring. These genes are known as linked genes (two different chromosomes), and they always have multiple alleles. Incompletely linked genes undergo crossing over, and the frequency of crossing over depends upon their distance. In contrast, the completely linked genes do not undergo crossing over at all. Means, they do not produce recombinant gametes. If two linked genes are separated by 10 cM (centi morgans or map units), the percent recombinants produced by this cross is always equal to the 10% and the remaining 90% will be identical to the parental genotype. And, in case of cross between the linked genes, the highest percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining are recombinants. Solution During meiosis, each member of a pair of genes tends to be randomly distributed into gametes (receive alleles) independently of how other chromosomes are distributed. Genes that are having their loci nearer to each other are not generally separated during chromosomal crossover and are inherited together to the offspring. These genes are known as linked genes (two different chromosomes), and they always have multiple alleles. Incompletely linked genes undergo crossing over, and the frequency of crossing over depends upon their distance. In contrast, the completely linked genes do not undergo crossing over at all. Means, they do not produce recombinant gametes. If two linked genes are separated by 10 cM (centi morgans or map units), the percent recombinants produced by this cross is always equal to the 10% and the remaining 90% will be identical to the parental genotype. And, in case of cross between the linked genes, the highest percent of offspring indicates (always, equal to or more than 50%) the paretnal and the remaining are recombinants..

During meiosis, each member of a pair of genes tends to be randomly .pdf

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B parents marital statusSolutionB parents marital status.pdf

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ANSWERS 12. B collecting ducts 13. B efferent arteriol 15. juxtaglomerular cells 16. A can be active or passive 17. C loop of Henle 18. D proximal convoluted tubule 19. .C increase the surface area of the mucosa of the small intestine 20. D submucosa of the duodenum 21. E myenteric plexus 22. D secretion - gall bladder epithelial cells ( I AM NOT SURE ABOUT QUESTION NO .14 , ) Solution ANSWERS 12. B collecting ducts 13. B efferent arteriol 15. juxtaglomerular cells 16. A can be active or passive 17. C loop of Henle 18. D proximal convoluted tubule 19. .C increase the surface area of the mucosa of the small intestine 20. D submucosa of the duodenum 21. E myenteric plexus 22. D secretion - gall bladder epithelial cells ( I AM NOT SURE ABOUT QUESTION NO .14 , ).

ANSWERS12. B collecting ducts13. B efferent arteriol15. juxtag.pdf

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Array:- Arrays is a collection of data items with same data type and access using a common name. Linked List:-A linked list is a linear data structure where each element is a separate object. Each element(node) of a list is comprising of two items - the data and a reference to the next node. Both Arrays and Linked Lists having advantages and disadvantages. Solution Array:- Arrays is a collection of data items with same data type and access using a common name. Linked List:-A linked list is a linear data structure where each element is a separate object. Each element(node) of a list is comprising of two items - the data and a reference to the next node. Both Arrays and Linked Lists having advantages and disadvantages..

Array- Arrays is a collection of data items with same data type and.pdf

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Ans. A. During protein folding, the hydrophobic residues are buried (placed) towards the core (center) of the proteins and the polar residues are preferentially exposed at the outer surface exposed to the aqueous environment of the cell. The hydrophobic residues interact among each other with hydrophobic interactions and van der Waals interactions. Hydrophobic interactions also exhibit positive cooperativity, that is, rate of protein folding increases as one hydrophobic interaction positively increases the occurrence of next hydrophobic interactions and so on. Moreover, the hydrophobic core stabilized by hydrophobic interactions in thermodynamically very stable. Ans. B. Air oxidation results the oxidation of thiol groups (-SH) of cysteine residues by removing their H-atoms, the resultant cysteine residues (oxidized) form disulfide bonds (-S-S) with other oxidized cysteine residues. -SH (of Cys residue 1) + -SH (of Cys residue 2) ---air oxidation--> -S-S- (disulfide bond) Since, the air oxidation is a random process, a denatured and reduced peptide with 10 Cys residues may form 5 disulfide bonds irrespective of the initial number of disulfide in the native protein conformation. Thus, number of disulfide bonds formed in (i) and (ii) is equal to 5. Solution Ans. A. During protein folding, the hydrophobic residues are buried (placed) towards the core (center) of the proteins and the polar residues are preferentially exposed at the outer surface exposed to the aqueous environment of the cell. The hydrophobic residues interact among each other with hydrophobic interactions and van der Waals interactions. Hydrophobic interactions also exhibit positive cooperativity, that is, rate of protein folding increases as one hydrophobic interaction positively increases the occurrence of next hydrophobic interactions and so on. Moreover, the hydrophobic core stabilized by hydrophobic interactions in thermodynamically very stable. Ans. B. Air oxidation results the oxidation of thiol groups (-SH) of cysteine residues by removing their H-atoms, the resultant cysteine residues (oxidized) form disulfide bonds (-S-S) with other oxidized cysteine residues. -SH (of Cys residue 1) + -SH (of Cys residue 2) ---air oxidation--> -S-S- (disulfide bond) Since, the air oxidation is a random process, a denatured and reduced peptide with 10 Cys residues may form 5 disulfide bonds irrespective of the initial number of disulfide in the native protein conformation. Thus, number of disulfide bonds formed in (i) and (ii) is equal to 5..

Ans. A. During protein folding, the hydrophobic residues are buried .pdf

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A) worst case complexity it will compare with all element of array so worst case time complexity = O(n) B) in this algorithm we will always compare the elements so its best and average case complexity is same as worst case complexity. Solution A) worst case complexity it will compare with all element of array so worst case time complexity = O(n) B) in this algorithm we will always compare the elements so its best and average case complexity is same as worst case complexity. .

A)worst case complexityit will compare with all element .pdf

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a) Our lungs are lined with the surfactant which has phospholipid and surfactant proteins. So as soon as the foreign silicon particle enters, it gets coated with the phospholipid and surfactant proteins. This changes the surface chemistry of silica and then it gets bind to alveolar macrophage. Macrophages too have certain receptors which play important role in this binding. b)When cell dies, it gets burst and release silica particles which again enter in endless cycle of infection. Solution a) Our lungs are lined with the surfactant which has phospholipid and surfactant proteins. So as soon as the foreign silicon particle enters, it gets coated with the phospholipid and surfactant proteins. This changes the surface chemistry of silica and then it gets bind to alveolar macrophage. Macrophages too have certain receptors which play important role in this binding. b)When cell dies, it gets burst and release silica particles which again enter in endless cycle of infection..

a) Our lungs are lined with the surfactant which has phospholipid an.pdf

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A four dimensional subspace cant be spanned by two vectors only. T.pdf

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12L has mass 16 grams 22.4L of gas has M mass M= .pdf

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2 m5275145Solution2 m5275145.pdf

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1. Personal Trainers Inc. is about to expand its’ successful fitness center into a new fitness “Super Center” in Toronto. This Super Center will include the necessities of all fitness centers, like a large exercise area with state of the art equipment, a swimming pool, a sporting goods store, a health food store, and a snack bar. Along with these, they will offer new aspects like child care, a teen center, and a computer cafe. 2. Account Recievable:Business recieves payment from customers for goods or services Account Payable:Payment that comes from the company to outside sources. Generel Ledger:This record keeps information on all transaction. Membership list/word processing:This is the information that will be kept on each member of the fitness center. 4. I believe Business Support would be a great asset for Personal Trainer Inc. to consider. For example, business support systems can keep track of what a company sells when working with at TP system. In this case, membership balances can be updated for each month they pay. Also, during the beginning of the process for creating the Super Center, a business support system can help users make decisions by creating a computer model and applying a set of variables. In this situation, they could use a what-if analysis to determine the price it must.. 5. Personal Trainer, Inc. owns and operates fitness centers in a dozen Midwestern cities. The centers Have done well, and the company is planning an international expansion by opening a new“Supercenter” in the Toronto area. Personal Trainer’s president, Cassia Umi, hired an ITConsultant, Susan Park, to help develop an information system for the new facility. During theProject, Susan will work closely with Gray Lewis, who will manage the new operation. Background: At their initial meeting, Susan and Gray discussed some initial steps in planning an information System for the new facility. The next morning, they worked together on a business profile, drew an Organization chart, discussed feasibility issues, and talked about various types of information Systems that would provide the best support for the supercenter’s operations. Their main objective Was to carry out a preliminary investigation of the new system and report their recommendations To Personal Trainer’s top managers. After the working session with Gray, Susan returned to her office and reviewed her notes. She Knew that Personal Trainer’s president, Cassia Umi, wanted the supercenter to become a model for The company’s future growth, but she did not remember any mention of an overall strategic planFor the company. Susan also wondered whether the firm had done a SWOT analysis or analyzedThe internal and external factors that might affect an information system for the supercenter.Because the new operation would be so important to the company, Susan believed thatPersonal Trainer should consider an enterprise resource planning strategy that could provide aCompany-wide framework for information man.

1.Personal Trainers Inc. is about to expand its’ successful fitnes.pdf

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