Write a java program to randomly generate the following sets of data.pdf

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Write a java program to randomly generate the following sets of data: 1.) 10 numbers 2.) 1,000 numbers 3.) 100,000 numbers 4.) 1,000,000 numbers 5.) 10,000,000 numbers Your program must sort the above sets of numbers using the following algorithms: 1.) Insertion Sort 2.) Merge Sort 3.) Quick Sort 4.) Heap Sort Print out the time each algorithm takes to sort the above numbers Solution import java.util.Random; public class Sort { private int[] array; private int size ; public Sort(int n){ array = new int[n]; size = n; Random rand = new Random(); for (int i = 0 ; i < n ; i++ ) { array[i] = rand.nextInt(100000) ; } } public void print(){ for (int i = 0; i < size ; i++ ) { System.out.print(array[i] + \" \"); } System.out.println(\"\"); } void insertion_sort() { for (int i=1; i=0 && array[j] > key) { array[j+1] = array[j]; j = j-1; } array[j+1] = key; } } void merge(int arr[], int l, int m, int r) { // Find sizes of two subarrays to be merged int n1 = m - l + 1; int n2 = r - m; /* Create temp arrays */ int L[] = new int [n1]; int R[] = new int [n2]; /*Copy data to temp arrays*/ for (int i=0; i Array to be sorted, low --> Starting index, high --> Ending index */ void quick_sort(int low, int high) { if (low < high) { /* pi is partitioning index, arr[pi] is now at right place */ int pi = partition(array, low, high); // Recursively sort elements before // partition and after partition quick_sort(low, pi-1); quick_sort(pi+1, high); } } public void heap_sort() { int n = array.length; // Build heap (rearrange array) for (int i = n / 2 - 1; i >= 0; i--) heapify(array, n, i); // One by one extract an element from heap for (int i=n-1; i>=0; i--) { // Move current root to end int temp = array[0]; array[0] = array[i]; array[i] = temp; // call max heapify on the reduced heap heapify(array, i, 0); } } // To heapify a subtree rooted with node i which is // an index in arr[]. n is size of heap void heapify(int arr[], int n, int i) { int largest = i; // Initialize largest as root int l = 2*i + 1; // left = 2*i + 1 int r = 2*i + 2; // right = 2*i + 2 // If left child is larger than root if (l < n && arr[l] > arr[largest]) largest = l; // If right child is larger than largest so far if (r < n && arr[r] > arr[largest]) largest = r; // If largest is not root if (largest != i) { int swap = arr[i]; arr[i] = arr[largest]; arr[largest] = swap; // Recursively heapify the affected sub-tree heapify(arr, n, largest); } } public static void main(String[] args) { Sort sort1 = new Sort(10); long start = System.currentTimeMillis(); sort1.insertion_sort(); long time = System.currentTimeMillis() - start; //sort1.print(); System.out.println(\"insertion_sort for 10 numbers take \" + time + \" ms \"); sort1 = new Sort(10); start = System.currentTimeMillis(); sort1.merge_sort(0,9); time = System.currentTimeMillis() - start; //sort1.print(); System.out.println(\"merge_sort for 10 numbers take \" + time + \" ms \"); sort1 = new Sort(10); start = System.currentTimeMillis(); sort1.quick_sor.

Education

$} void insertion_sort() { for (int i=1; i=0 && array[j] > key) { array[j+1] = array[j]; j = j-1; } array[j+1] = key; } } void merge(int arr[], int l, int m, int r) { // Find sizes of two subarrays to be merged int n1 = m - l + 1; int n2 = r - m; /* Create temp arrays */ int L[] = new int [n1]; int R[] = new int [n2]; /*Copy data to temp arrays*/ for (int i=0; i Array to be sorted, low --> Starting index, high --> Ending index */ void quick_sort(int low, int high) { if (low < high) { /* pi is partitioning index, arr[pi] is now at right place */ int pi = partition(array, low, high); // Recursively sort elements before // partition and after partition$

$quick_sort(low, pi-1); quick_sort(pi+1, high); } } public void heap_sort() { int n = array.length; // Build heap (rearrange array) for (int i = n / 2 - 1; i >= 0; i--) heapify(array, n, i); // One by one extract an element from heap for (int i=n-1; i>=0; i--) { // Move current root to end int temp = array[0]; array[0] = array[i]; array[i] = temp; // call max heapify on the reduced heap heapify(array, i, 0); } } // To heapify a subtree rooted with node i which is // an index in arr[]. n is size of heap void heapify(int arr[], int n, int i) { int largest = i; // Initialize largest as root int l = 2*i + 1; // left = 2*i + 1 int r = 2*i + 2; // right = 2*i + 2 // If left child is larger than root if (l < n && arr[l] > arr[largest])$

$largest = l; // If right child is larger than largest so far if (r < n && arr[r] > arr[largest]) largest = r; // If largest is not root if (largest != i) { int swap = arr[i]; arr[i] = arr[largest]; arr[largest] = swap; // Recursively heapify the affected sub-tree heapify(arr, n, largest); } } public static void main(String[] args) { Sort sort1 = new Sort(10); long start = System.currentTimeMillis(); sort1.insertion_sort(); long time = System.currentTimeMillis() - start; //sort1.print(); System.out.println("insertion_sort for 10 numbers take " + time + " ms "); sort1 = new Sort(10); start = System.currentTimeMillis(); sort1.merge_sort(0,9); time = System.currentTimeMillis() - start; //sort1.print(); System.out.println("merge_sort for 10 numbers take " + time + " ms "); sort1 = new Sort(10); start = System.currentTimeMillis(); sort1.quick_sort(0,9); time = System.currentTimeMillis() - start; //sort1.print();$

System.out.println("quick_sort for 10 numbers take " + time + " ms ");
sort1 = new Sort(10);
start = System.currentTimeMillis();
sort1.heap_sort();
time = System.currentTimeMillis() - start;
//sort1.print();
System.out.println("heap_sort for 10 numbers take " + time + " ms ");
////////////////////////////////////////////////////////////////////////
System.out.println("");System.out.println("");
sort1 = new Sort(1000);
start = System.currentTimeMillis();
sort1.insertion_sort();
time = System.currentTimeMillis() - start;
System.out.println("insertion_sort for 1000 numbers take " + time + " ms ");
sort1 = new Sort(1000);
start = System.currentTimeMillis();
sort1.merge_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("merge_sort for 1000 numbers take " + time + " ms ");
sort1 = new Sort(1000);
start = System.currentTimeMillis();
sort1.quick_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("quick_sort for 1000 numbers take " + time + " ms ");
sort1 = new Sort(1000);
start = System.currentTimeMillis();
sort1.heap_sort();
time = System.currentTimeMillis() - start;
System.out.println("heap_sort for 1000 numbers take " + time + " ms ");

////////////////////////////////////////////////////////////////////////
System.out.println("");System.out.println("");
sort1 = new Sort(100000);
start = System.currentTimeMillis();
sort1.insertion_sort();
time = System.currentTimeMillis() - start;
System.out.println("insertion_sort for 100000 numbers take " + time + " ms ");
sort1 = new Sort(100000);
start = System.currentTimeMillis();
sort1.merge_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("merge_sort for 100000 numbers take " + time + " ms ");
sort1 = new Sort(100000);
start = System.currentTimeMillis();
sort1.quick_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("quick_sort for 100000 numbers take " + time + " ms ");
sort1 = new Sort(100000);
start = System.currentTimeMillis();
sort1.heap_sort();
time = System.currentTimeMillis() - start;
System.out.println("heap sort for 100000 numbers take " + time + " ms ");
////////////////////////////////////////////////////////////////////////
System.out.println("");System.out.println("");
sort1 = new Sort(1000000);
start = System.currentTimeMillis();
sort1.insertion_sort();
time = System.currentTimeMillis() - start;

System.out.println("insertion_sort for 1000000 numbers take " + time + " ms ");
sort1 = new Sort(1000000);
start = System.currentTimeMillis();
sort1.merge_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("merge_sort for 1000000 numbers take " + time + " ms ");
sort1 = new Sort(1000000);
start = System.currentTimeMillis();
sort1.quick_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("quick_sort for 1000000 numbers take " + time + " ms ");
sort1 = new Sort(1000000);
start = System.currentTimeMillis();
sort1.heap_sort();
time = System.currentTimeMillis() - start;
System.out.println("heap_sort for 1000000 numbers take " + time + " ms ");
////////////////////////////////////////////////////////////////////////
System.out.println("");System.out.println("");
sort1 = new Sort(10000000);
start = System.currentTimeMillis();
sort1.insertion_sort();
time = System.currentTimeMillis() - start;
System.out.println("insertion_sort for 10000000 numbers take " + time + " ms ");
sort1 = new Sort(10000000);
start = System.currentTimeMillis();
sort1.merge_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("merge_sort for 10000000 numbers take " + time + " ms ");
sort1 = new Sort(10000000);

start = System.currentTimeMillis();
sort1.quick_sort(0,9);
time = System.currentTimeMillis() - start;
System.out.println("quick_sort for 10000000 numbers take " + time + " ms ");
sort1 = new Sort(10000000);
start = System.currentTimeMillis();
sort1.heap_sort();
time = System.currentTimeMillis() - start;
}
}
Sample Output
insertion_sort for 10 numbers take 0 ms
merge_sort for 10 numbers take 0 ms
quick_sort for 10 numbers take 0 ms
heap_sort for 10 numbers take 0 ms
insertion_sort for 1000 numbers take 5 ms
merge_sort for 1000 numbers take 0 ms
quick_sort for 1000 numbers take 0 ms
heap_sort for 1000 numbers take 6 ms
insertion_sort for 100000 numbers take 1239 ms
merge_sort for 100000 numbers take 0 ms
quick_sort for 100000 numbers take 0 ms
heap sort for 100000 numbers take 21 ms
insertion_sort for 1000000 numbers take 217971 ms
merge_sort for 1000000 numbers take 0 ms
quick_sort for 1000000 numbers take 0 ms
heap_sort for 1000000 numbers take 261 ms

Similar to Write a java program to randomly generate the following sets of data.pdf

import java.util.Arrays; import java.io.PrintWriter; import java.io.File; import java.util.Scanner; public class SortTimes { /* convert nanos to seconds */ static final double ONE_BILLION = 1_000_000_000.; /* sizes of input range should range up to 512k elements */ static int[] sizes = new int[32]; public static void main(String[] args) throws Exception { int sIdx = 0; long startTime; long endTime; double time; int size = 128*1024/sizes.length; /* let's make size smaller, comment out this next line and see what happens */ size = 16*1024/sizes.length; for( int i=0; i<sizes.length; i++ ) { sizes[i] = (i+1)*size; } Scanner inp = new Scanner(System.in); System.out.print("What is your last name? "); String lastName = inp.next(); PrintWriter pw = new PrintWriter(new File(lastName + "_sortTimes.csv")); pw.println( "Size, BubbleSort, MergeSort, QuickSort, JavaSort, BubbleSort (Sorted Array), MergeSort (Sorted Array), QuickSort (Sorted Array), JavaSort (Sorted Array)" ); while( sIdx < sizes.length ) { int[] A = new int[sizes[sIdx]]; int[] B = new int[sizes[sIdx]]; int[] C = new int[sizes[sIdx]]; int[] D = new int[sizes[sIdx]]; int[] E = new int[sizes[sIdx]]; for( int i=0; i<A.length; i++ ) { A[i] = (int)(Math.random()*sizes[sIdx]*2); B[i] = A[i]; C[i] = A[i]; D[i] = A[i]; E[i] = 1000; // Array E is used for testing algo on array that is already sorted. } System.out.println(); System.out.println("--------------------------------------------------"); System.out.println("Initial Values: "); printArray(A); pw.print(A.length + ", "); System.out.println( "Starting bubble sort ---------------- Size = " + B.length); startTime = System.nanoTime(); bubbleSort(A); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Bubble sort finished - time = " + time + " seconds" ); pw.print(time + ", "); printArray(A); System.out.println(); System.out.println( "Starting merge sort ---------------- Size = " + C.length); startTime = System.nanoTime(); mergeSort(B); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Merge sort finished - time = " + time + " seconds."); pw.print(time + ", "); printArray(B); System.out.println(); System.out.println( "Starting quick sort ---------------- Size = " + D.length); startTime = System.nanoTime(); quickSort(C); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "quickSort finished - time = " + time + " seconds." ); pw.print(time + ", "); printArray(C); System.out.println(); System.out.println( "Starting java's Array sort ---------------- Size = " + D.length); startTime = System.nanoTime(); javaSort(D); endTime = System.nanoTime(); time = (endTime - startTime) / ONE_BILLION; System.out.println( "Java's Array sort finished - time = " + time + " seconds." ); pw.print(time + ", "); printArray(D); System.out.println(); System.out.println( "Starting bubble sort on sorted array ---------------- Size = " + B.length); star.

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Can you give an example of a binary heap program Can you give an example of a binary heap program Solution /* * C Program to Implement a Heap & provide Insertion & Deletion Operation */ #include int array[100], n; main() { int choice, num; n = 0;/*Represents number of nodes in the heap*/ while(1) { printf(\"1.Insert the element \ \"); printf(\"2.Delete the element \ \"); printf(\"3.Display all elements \ \"); printf(\"4.Quit \ \"); printf(\"Enter your choice : \"); scanf(\"%d\", &choice); switch(choice) { case 1: printf(\"Enter the element to be inserted to the list : \"); scanf(\"%d\", &num); insert(num, n); n = n + 1; break; case 2: printf(\"Enter the elements to be deleted from the list: \"); scanf(\"%d\", &num); delete(num); break; case 3: display(); break; case 4: exit(0); default: printf(\"Invalid choice \ \"); }/*End of switch */ }/*End of while */ }/*End of main()*/ display() { int i; if (n == 0) { printf(\"Heap is empty \ \"); return; } for (i = 0; i < n; i++) printf(\"%d \", array[i]); printf(\"\ \"); }/*End of display()*/ insert(int num, int location) { int parentnode; while (location > 0) { parentnode =(location - 1)/2; if (num <= array[parentnode]) { array[location] = num; return; } array[location] = array[parentnode]; location = parentnode; }/*End of while*/ array[0] = num; /*assign number to the root node */ }/*End of insert()*/ delete(int num) { int left, right, i, temp, parentnode; for (i = 0; i < num; i++) { if (num == array[i]) break; } if (num != array[i]) { printf(\"%d not found in heap list\ \", num); return; } array[i] = array[n - 1]; n = n - 1; parentnode =(i - 1) / 2; /*find parentnode of node i */ if (array[i] > array[parentnode]) { insert(array[i], i); return; } left = 2 * i + 1; /*left child of i*/ right = 2 * i + 2; /* right child of i*/ while (right < n) { if (array[i] >= array[left] && array[i] >= array[right]) return; if (array[right] <= array[left]) { temp = array[i]; array[i] = array[left]; array[left] = temp; i = left; } else { temp = array[i]; array[i] = array[right]; array[right] = temp; i = right; } left = 2 * i + 1; right = 2 * i + 2; }/*End of while*/ if (left == n - 1 && array[i]) { temp = array[i]; array[i] = array[left]; array[left] = temp; } } Output:.

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Bats rely on sound localization to be able to fly during the night. On a cool 20 degree C summer night, a bat produces an ultrasonic sound with a frequency of 37 kHz. What is the wavelength of this ultrasound, in meters? While remaining stationary, the bat emits a sound pulse and receives an echo after 0.130 s. How far away, in meters, is the reflecting object from the bat? Solution 1) wave length of ultra sound waves = velocity / frequency = 330/37kHz ( velocity of ultra sound waves in air = 330 m/sec) =.00089189m 2) time taken = .130 distance travelled for the sound pulse = 2*d ( because it is also returing back after strucking the object) v*t=2*d d= 330*.130/2=21.45m.

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Discuss the impact that the tonnage of sulfur or sulfuric acid recovered from SO2 in flue gas treatment would have upon the chemical industry if all coal- and oil-fired were required to use such treatment processes. Solution YEs, definitely we must required to use FGD as sulfur dioxide enters the air mainly from industrial processes and from the combustion of hydrocarbon fuels with a substantial sulfur (S) content and represents a major source of air pollution. Sulfur dioxide is a colorless, corrosive gas that has a bitter taste, but no smell at low levels. The share of SO2 emissions comes from the use of coal and oil in fossil-fired power plants, in industrial combustion units, in small combustion units in households, and in vehicles. Besides the concentration of SO2 in stack gases emitted by steam generation plants is usually in the range of 800 to 6000 mg/Nm3, the volume of gases produced by the utility industry worldwide results in the liberation of a large tonnage of sulfur dioxide into the atmosphere.Sulfur dioxide is a chemical that can be dangerous in many ways. At lower levels, sulfur dioxide has been found to be a corrosive irritant to eyes and skin. Sulfur dioxide has been associated with a variety of respiratory diseases and increased mortality rates. Inhalation of sulfur dioxide can cause increased airway resistance by constricting lung passages. Sulfur dioxide is also one of the main ingredients in acid rain. Acid rain occurs when sulfur dioxide or other gaseous chemicals, such as nitrogen oxides, are released into the air. These gases ascend through the atmosphere, and when they reach upper cloud levels, they react with water, oxygen, and sunlight.Various concentrations of sulfuric acid and nitric acid are then produced. These acids mix with the condensed water vapor in the clouds and fall to the ground with the water in various forms of precipitation. This precipitation with greater acidity is what is known as acid rain. Acid rain is dangerous to many natural ecosystems; it can harm plants, sand, make water undrinkable, and unlivable for many animals. The biggest producers of sulfur dioxide emissions are electrical utilities that produce electricity through the burning of fossil fuels.Sulfur dioxide emissions, from combustion installations using coal or HFO, can be reduced by fuel desulfurization or by FGD, as well as by a combination of these two measures. Flue gas desulfurization is an efficient method for the reduction of the sulfur dioxide emissions.2 Many processes are available in the market, such as (a) wet scrubbers, (b) spray dry scrubbers, (c) sorbent injection, (d) regenerable processes, and (e) combined SO2/NOX removal processes.

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You have decided to go to graduate school for Molecular Biology (yay!). Your first project involves investigating the regulation of expression of a gene suspected to be involved in the immune response to Zika Virus (we\'ll call this gene ZV1). You have found that this gene is expressed when someone is infected by Zika, but is not expressed in uninfected people. You believe that this is due to a change in a histone modification. Come up with a hypothesis for what type of change in histone modification may be responsible for the expression of this gene upon infection with Zika. Describe, in detail, an experiment to test your hypothesis. Include any necessary controls. Assuming your hypothesis is correct (of course it is), draw the results of the experiment (it doesn\'t have to be perfect, do the best you can). One of your friends in graduate school is super interested in your project. She thinks that there may be an epigenetic component to the expression of this gene in response to Zika Virus other than that of histone modifications. Describe a non-chromatin based mechanism for this and briefly describe how you might test this. Solution Thanks for the question. a. A hypothesis for the mechanism of expression of gene following infection with zika virus is follows the zika virus upon entering the body produces a translation whereby a RNA genome is produced which enters the endosomal membrane of the host cell causing the symptoms of zika virus fever etc b. an elisa test (enzyme linked immuno sorbent assay) which could test either the zika virus or its translated RNA could be a good way of assessing the infection of zika virus. c. A standard can be set up on the number of zika virus or its replicated RNA which is necessary to be present in the body of the infected patient to call him to be actively infected with the virus. This will decide for subsequent patients if they are infected with the virus and also if they are actively carrying the infection. d. the original hypothesis stated above is already having an epigenetic component. Hence its detection and other procedures are already mentioned above. Regards.

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Why was any carbon-containing molecule originally called \"organic\"? A. Carbon-containing molecules were thought to be only produced by organisms. B. Carbon-containing molecules contain organic substances within them that cannot be synthesized inorganically. C. Carbon- containing molecules do not contain hydrogen or oxygen, the inorganic parts of organisms. D. Carbon-containing molecules were thought to be present only outside of organisms - in the inorganic areas of the earth: water, air, and stone. Solution Correct answer (A) It was once thought that organic compounds could only be created by living organisms..

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Which of the following is true about indexes and index design? a. indexes usually speed up write operations b. a composite index is an index on a single column c. indexes should be created on columns used by WHERE and BY clauses d. an index is created for every foreign key by default Solution Answer: Indexes should be created on columns used by WHERE and BY clauses. Indexes are used for retrieve the records fast. So we have to mention the column names in WHERE clause and By clause in which we created the columns for fast retrieval of the records..

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What kind of organisms likely perished due to the increase of oxygen in the atmosphere three billion years ago? heterotrophs cyanobacteria obligate aerobes obligate anaerobes facultative anaerobes In the accompanying figure, note the variation in these closely related birds. What does this illustrate\'? hybridization macroevolution hybrid inviability adaptive radiation allometric radiation One critical requirement for life to arise on Earth was time for which to occur? The salinity of the oceans to stabilize The Earth\'s various climates to stabilize Molecules to accumulate and react with one another Solution 6 . B cyanobacteria Cyanobacteria increased in high amount 3 billion yrs ago dur to sudden increase in oxygen in the environment which caused fires. Anaerobes couldn\'t survive, as oxygen was toxic for them.

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what is the \"HTTP\" protocol used for? What are its major parts? Solution it is a application protocol for data communication for the world wide web request and response method: it contains what type of action does client want to perform uri:it is know as uniform resource indicator it will have the address of the required resource header fileds:it will tell extra information about the request body:it contains the data which sent by the client to the server via: these are used proxies to display intermediate sites that the request passes.

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What is the between facial action unit and facial landmark differences ? Solution There is a difference between the facial action unit and facial landmark. The difference can be illustrated as follows: FACIAL ACTION UNIT: Humans interact far more naturally with each other tahn they do with machines. To approach the naturalness of face-to-face human interaction machines should be able to emulate the way humans communicate with each other. Although speech alone is often sufficient for communicating with another person, non-verbal communicative cues can help to synchronize the dialogue, to signal comprehensionor disagreement and to let the dialogue run smoother, with less interruptions. With facial expresssions we clarify what is said by means of lip-reading, we stress the importance of the spoken message by means of conversational signals like raisisng eyebrows, and siognal comprehension,disagrreement,boredom and intensions. Machine understanding of facial expressions could revolutionize user interfaces including ambient,automotiveand robot interfaces and has become,therefore,a hot topic in computer-vision approach. FACIAL LANDMARK: Localizing facial landmarks (i.e facial alignment) is a fundamental step in facial image analysis. However, the problem is still challenging due to the large variability in pose and appearence, and the existence of occlusions in real-world face images. EGM (Exemplar-based graph matching) framework is used in the facial landmark localization.Compared to conventional algorithms, EGM has three advantages:.

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What are contemporary management challenges and opportunities? Solution Contemporary management challenges are:- The character of ownership Globalization and anti-globalization Social sustainability Corporate social responsibility and corporate criminality Environmentalsustainability Organizational misbehaviour and resistance Corporate governance Masculinisation and Feminisation of Management Organizational identity and identity work New forms of work such as emotional labour and aesthetic labour New organizational forms.

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Valinomycin is a potassium ionophore. What would be its effect on insulin secretion by cells that were exposed to valinomycin? Solution Ionosphore are the molecules which can transfer a particular ions across a semi-permeambke membrane..!! Valinomycin is also a ionophore which transfer the potassium across the cell memberane.. Due to which when it is exposed to the beta cells of insulin it inhibits the secretion of insulin in the body as potassium plays a major role in insulin secretion.. As the concentration of potassium is disturbed in the beta cells in inhibits the insulin secretion.. Hence the blood glucose level increases..!!.

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The retrovirus genome is made of the nucleic acid before the virus becomes a provirus, in order for that to occur the form of the genome must be converted to Solution 1) Answer: The retrovirus genome is made of the nucleic acid, RNA before the virus become a provirus; in order that to occur, single stranded form of the genome must be converted to double stranded form. Explanation: A retrovirus is a single-stranded RNA virus with a DNA intermediate and, as an obligate parasite, targets a host cell. In most viruses, DNA is transcribed into RNA, and then RNA is translated into protein. However, retroviruses function differently, as their RNA is reverse-transcribed into DNA (by an enzyme reverse transcriptase), which is integrated into the host cell\'s genome (when it becomes a provirus), and then undergoes the usual transcription and translational processes to express the genes carried by the virus. 2) Answer: Human make RNA during a process called transcription. Synthesis of RNA is usually catalyzed by an enzyme—RNA polymerase—using DNA as a template, a process known as transcription..

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the probability that it is raining is .25. the orbability that it is windy is .30. if it is windy, the probability that it is raining is .40. what is the probabilty that it is raining and windy? assume that the events raining and windy are not independent. Solution Let R denote the event of raining and W that of winding. Given, P(R)=.25 P(W)=.3 and P(R|W) = .4 We know P(R|W) = P(RW) /P(W) => P(RW) = P(R|W)*P(W) =>P(RW)=P(It is raining and windy)=(.4)*(.3)=0.12.

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Significance of Discoveries in Genetics and DNA Our understanding of genetic inheritance and the function of DNA in producing the characteristics of the individual have been developing for more than 150 years. Consider our current state of knowledge. Link genetic characteristics to DNA structure. Explain how DNA through the process of protein synthesis is responsible for the ultimate expression of the characteristics in the organism. Describe how interference in protein synthesis can result in disruption of cellular and bodily processes? How does the significance of one class of proteins, the enzymes, relate? Please answer all questions! Solution 1. Genes are segments of chromosome and correspond to DNA sequence(s). The DNA segment carries coded information for the synthesis of specific proteins which carry out different functions in the cell. The characters are expressed through proteins. DNA is packed into chromosomes. In gametes, the chromosome number is halved. The fusion of gametes in zygotes restores the chromosome number. 2. Segments of DNA called genes carry traits in encoded form. The genes undergo transcription (transfer of information from DNA to RNA). This is followed by translation which is the transfer of information from the RNA to the protein. The characters are expressed through proteins which are specific in function. 3. Various antimicrobial drugs interfere with the process of protein synthesis at the ribosome level. The information encoded in the nucleotides of DNA are expressed through proteins which carry out specific functions in the cell. Disruption of protein synthesis prevents the biochemical pathways and specific functions of proteins. 4. Enzymes are responsible for controlling various vital activities in the cell. They synthesize, deliver and eliminate various chemicals. Enzymes undergo slow depletion which reduces overall strength and viatality of the human body..

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Scope is much less complicated if functions cannot contain other function definitions (C++, Java: cannot; Ada, Pascal, Ada, Python, Ruby, Lisp: can). Give one good reason why functions should not contain other function definitions. Give one good reason why it should be allowed. Solution A function is the smallest block of code to perform a specific task. Which can be independently invoked. The variable defined in a function are available only to that block not to the whole program which gives abstraction. the scope and life of variable defined in the funciton are available only when it is invoked, which saves memory as well. b. To keep a function isolated. It\'s only used inside the function which is invoked i.e. one place. when it\'s locally used why define globally.It\'s just another way of breaking down a big function into smaller pieces without affecting the global namespace with another function name. Quite often the inner function isn\'t a stand-alone so doesn\'t rightfully belong in the global namespace..

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Refer to the drawings in Figure 13.2 of a single pair of homologous chromosomes as they might appear during various stages of either mitosis or meiosis, and answer the following questions (28). Which diagram represents chromosomes in prophase I of meiosis? I II III IV V VI Which diagram represents anaphase II of meiosis? I III IV V Solution 26) IV Explanation: During prophase I of meiosis, synapsed homologous chromosomes are observed as tetrads. 27) II Explanation: There is separation of centromeres. As a result, chromatids separate and they are now called independent chromosomes..

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