Answer True or False to the following statements Colony is forme.pdf

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Answer \"True or False\" to the following statements Colony is formed from a large number of bacterial cells. Both spread-plate and pour plate methods will only yield pure culture. Turbidometeric methods count only live cells. Increase in turbidity is an index of bacterial growth and biomass. The primary stain of the gram\'s staining reaction is called safranin. The secondary/counter stain of the grams staining reaction is methylene blue. Gram positive cells have an extra layer in the cell component. Carbolfuchsin is used in staining acid fast organisms. Nucleic acids usually bind to basic dyes. Blood agar medium is an enrichment medium instead of a selective medium. A clone consists different kinds of bacteria. A pure culture can be formed from streaked plate method. Solution Group of bacteria grow on solid media called as colony. Hence colony formed by large number of bacterial cells. True statement. Spread plate method gives the pure culture. Pour plating dose not gives the pure culture , used for quantitative urine cultures.meny methods are there for isolation of pure culture example, enrichment media, selective media, indicator media, pretreatment, inoculation etc. Hence false statement. Turbidometry counts both live and dead cells. Hence false statement. True statement, increased turbidity index for bacterial growth and biomass. False, safranine is used for counter staining ,but not for primary staining. False, methylene blue is used for primary staining not for secondary staining. False statement, cell wall of positive bacteria has peptidoglycon layer and techoic acid. Gram negative has, lipopolysaccharides, outer membrane, lipoproteins and peptidoglycon. True statement, carbolfuchsin used for staining acid fast organisms. True statement, nucleic acids bind to basic dyes. True statement, blood agar is not selective media. False statement, clone consis of similar bacteria. True statement, streak culture or surface plating is used in pure culture..

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2. External events include all of the following except Purchasing equ.pdf

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The Missequa Valley Protectors understand groundwater is a great dri to understand how the water gets into each of their aquifers, so they can work to protect the supplies for generations to come. Compare and contrast how the unconsolidated, unconfined (UU) and the consolidated, confined (CC) aquifers are recharged. Make at least two of your own assumptions concerning topography of the area and presence of surface waters, which can aid discussion. Compare and contrast threats to both the UU and the CC aquifers, where at least two different scenarios are discussed (e.g. pollution, population change, land use changes, climate change, ..). ompare and otrast hredeat Solution recharge methods of the unconsolidated unconfined aquifer are ground seepage water and shallow wells while recharge methods of the consolidated confined aquifer are rock fractures, springs, and deep wells. main threats to unconsolidated unconfined aquifer are pollution and population change while for consolidated confined aquifer its population, land use change, deforestation and climate change..

The Missequa Valley Protectors understand groundwater is a great dri .pdf

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The If and IF else are a pair of commands in the Arduino Processing language that handle: Solution The If and IF else are a pair of commands in the Arduino Processing language that handle: IF (conditional) and ==, !=, <, > (comparison operators/functions) IF function, which is used in conjunction with a comparison operator, to tests whether a certain condition has been reached or matched, such as an input being above a certain number. The format for an if test is: x = 70 IF (variable > x ) IF else.

The If and IF else are a pair of commands in the Arduino Processing .pdf

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The duct from an expanded chamber called the node truck A substance that provides an response is called a (n) complement T is so _ as B is to _. Bottom None of the answers is correct. Which of these statements about lymph flow in a node is false? Lymph enters in an afferent Lymph first enters the space. Lymph flows past cells. The efferent exits at the helium. The altered lymphatic enters at the helium. _ are Large lymphoid modules that are located in the walls of the pharynx. Tonsils Payer\'s patches Lymph nodes complexes Spleens Solution 24. E) cisterna chyli. It is a dilated sac at the lower end of the thoracic duct. The lymph from intestinal tuck and two lumbar lymphatic trunks flow into the cisterna chlyli. 25. E) Antigen. Antigen can induce immune response because it is treated a foreign body and our cells respond to it by producing antibodies against it. 26. B) thymus dependent, bone marrow derived The T –cells are matured in Thymus, so called as Thymus dependent. The B- cells are produced and matured in bone marrow . 27. E) The afferent lymphatics enter at the hilum. 28. E) Tonsils. It is a small lymphoid tissue that is located in the lateral walls of the opening between the mouth and the pharynx.

The duct from an expanded chamber called the node truck A substance t.pdf

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Suppose you have a method triPartition, whose documentation is given as follows: /* triPartition: Given TWO pivots, pivot1 and pivot2, * partition an unsorted list into THREE pieces. * * Details: Partition unsorted[start]...unsorted[end-1] into * <--smalls--><--mediums--><--bigs--> * where smalls = elements < pivot1 (in no particular order) * = unsorted[start]...unsorted[pivot1Posn] * mediums = elements < pivot2 and >= pivot1 (in no * particular order). * = unsorted[pivot1Posn+1]...unsorted[pivot2Posn] * bigs = elements >= pivot2 (in no particular order) * = unsorted[pivot2Posn+1]...unsorted[end-1] * pivot1 is in its correct position in the sorted list * = unsorted[pivot1Posn]. * pivot2 is in its correct position in the sorted list * = unsorted[pivot2Posn]. * assumes: start + 2 <= end (unsorted has at least two elements) * @param unsorted * IN: Array unsorted contains the integers to be partitioned. * pivot1 and pivot2 are already in unsorted[start] and * unsorted[start+1] (note that pivot1 <= pivot2). * OUT: unsorted is partitioned into * <--smalls--><--mediums--><--BIGS--> * @param start * Index in unsorted defining the sublist to be partitioned. * unsorted[start] ... unsorted[end-1] is to be partitioned. * @param end * Index in unsorted defining the sublist to be partitioned. * unsorted[start] ... unsorted[end-1] is to be partitioned. * @returns an object of type Pair, with two public (int) fields: * - pivot1Posn : the position of pivot1 * - pivot2Posn : the position of pivot2 */ The code for the Pair class is: public class Pair { public int pivot1Posn; public int pivot2Posn; } QUESTION: Write a recursive quicksort method that uses the method triPartition to sort an array of ints. You may assume the existence of a swap method, if necessary. Solution public static void QuickSort_Recursive(int[] arr, int left, int right) { // For Recusrion if(left < right) { int pivot = Partition(arr, left, right); if(pivot > 1) QuickSort_Recursive(arr, left, pivot - 1); if(pivot + 1 < right) QuickSort_Recursive(arr, pivot + 1, right); } } public static void main(String[] args) { int[] numbers = { 3, 8, 7, 5, 2, 1, 9, 6, 4 }; int len = 9; System.out.println(\"QuickSort By Recursive Method\"); QuickSort_Recursive(numbers, 0, len - 1); for (int i = 0; i < 9; i++) System.out.println(numbers[i]); System.out.println(); } }.

Suppose you have a method triPartition, whose documentation is given.pdf

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Task: Write a Java program to implement a simple graphics editor that can be used to draw a Pythagoras. The editor has a pull-down menu on top of the screen, with 2 buttons: \"Pythagoras\" and \"Quit\". 1) When the user selects \"Pythagoras\", he/she can draw a tree of Pythagoras on the screen. 2) The editor terminates/quits execution if the user selects \"Quit\" from the pulldown menu. 3) The editor has a horizontal and a vertical sliding bar that when slided, moves the tree on the canvas accordingly. Solution HOpe this will help-- import java.awt.event.ActionEvent; import java.awt.event.ActionListener; import javax.swing.ButtonGroup; import javax.swing.JFrame; import javax.swing.JMenu; import javax.swing.JMenuBar; import javax.swing.JMenuItem; public class MenuExp extends JFrame { public MenuExp() { setTitle(\"Menu Example\"); setSize(150, 150); JSlider framesPerSecond = new JSlider(JSlider.VERTICAL,FPS_MIN, FPS_MAX, FPS_INIT); framesPerSecond.addChangeListener(this); JSlider framesPerSecond1= new JSlider(JSlider.HORIZONTAL,FPS_MIN, FPS_MAX, FPS_INIT); framesPerSecond1.addChangeListener(this); // Creates a menubar for a JFrame JMenuBar menuBar = new JMenuBar(); // Add the menubar to the frame setJMenuBar(menuBar); // Define and add two drop down menu to the menubar JMenu fileMenu = new JMenu(\"Process\"); menuBar.add(fileMenu); // Create and add simple menu item to one of the drop down menu JMenuItem newAction = new JMenuItem(\"Pythagoras\"); JMenuItem openAction = new JMenuItem(\"Quit\"); fileMenu.add(newAction); fileMenu.add(openAction); newAction.addActionListner(new PythagorusListener()); openAction.addActionListner(new CloseListener()); // Add a listener to the New menu item. actionPerformed() method will // invoked, if user triggred this menu item newAction.addActionListener(new ActionListener() { public void actionPerformed(ActionEvent arg0) { System.out.println(\"You have clicked on the new action\"); } }); } public static void main(String[] args) { MenuExp me = new MenuExp(); me.setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); me.setVisible(true); } } public class CloseListener implements ActionListener{ @Override public void actionPerformed(ActionEvent e) { //DO SOMETHING System.exit(0); } } public class PythagorusListener implements ActionListener{ @Override public void actionPerformed(ActionEvent e) { setSize(900, 900); setTitle(\"Pythagoras tree\"); add(new Draw(n)); setDefaultCloseOperation(JFrame.EXIT_ON_CLOSE); setVisible(true); } } class Draw extends JComponent { private int height = 800; private int width = 800; private int steps; public Draw(int n) { steps = n; Dimension d = new Dimension(width, height); setMinimumSize(d); setPreferredSize(d); setMaximumSize(d); } @Override public void paintComponent(Graphics g) { super.paintComponent(g); g.setColor(Color.white); g.fillRect(0, 0, width, height); g.setColor(Color.black); int x1, x2, x3, y1, y2, y3; int base = width/7; x1 = (width/2)-(base/2); x2 = (width/2)+(base/2); x3 = w.

Task Write a Java program to implement a simple graphics editor tha.pdf

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A CALLING MODEM A calling modem transmits each data bit 1 as a 1270 hertz tone lasting one time unit, and each data bit 0 as a 1070 hertz tone . Write a program that displays and writes to an output file messages indicating the tones that would be emitted for the input data , a file of zeros and ones separated by spaces. The messages should take the form: Emit____ -hz tone for ___________ time unit(s). Be sure to prompt the user for file names. input file: 1 0 0 0 0 1 1 0 1 0 1 1 1 1 0 0 0 1 output: Emit 1270-hz tone for 1 time unit(s). Emit 1070-hz tone for 4 time units(s). Emit 1270-hr tone for 2 time units(s). Solution //This is a C code. It takes an \"output.txt\" as input. #include int main(void) { FILE *fp; char c; int i,count1=0,count0=0; fp = fopen(\"output.txt\", \"r\"); // error check this! while((c = fgetc(fp)) != EOF) { if (c == \'1\') { if(count0!=0) { printf(\"Emit 1070-hz tone for %d time unit(s).\ \",count0); count0=0; } count1++; } if(c == \'0\') { if(count1!=0) { printf(\"Emit 1270-hz tone for %d time unit(s).\ \",count1); count1=0; } count0++; } } if(count1!=0) { printf(\"Emit 1270-hz tone for %d time unit(s).\ \",count1); count1=0; } if(count0!=0) { printf(\"Emit 1070-hz tone for %d time unit(s).\ \",count0); count0=0; } fclose(fp); return 0; }.

A CALLING MODEMA calling modem transmits each data bit 1 as a 1270.pdf

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Show that L = {a^k | k is a square} is not regular. Solution Let p be the pumping length for the language. Let s = a^(p^2), according to pumping lemma s can be divided in to three parts xyz such that 1) |xy|<=P 2) |y| > 0 3) xy^iz belongs to L. As |xy|<=p which implies |y|<=p lets assumes y = a^j where j<=p, lets consider xy^2z which will be a^(p2+j). p^2 + j <= p^2 + p < p^2 + 2p + 1 < (p+1)^2. Hence xy^2z does not belong to L. Therefore it is not a regular language..

$Show that L = {a^k k is a square} is not regular.SolutionLet .pdf$ $Show that L = {a^k k is a square} is not regular.SolutionLet .pdf$

Show that L = {a^k k is a square} is not regular.SolutionLet .pdf

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