Albinism in humans is sometimes caused by a mutation in the OCA2 gene.pdf

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Albinism in humans is sometimes caused by a mutation in the OCA2 gene. Albinism is a __ while the mutant DNA sequence in the OCA2 gene that causes albinism is a genome; allele allele; chromosome mutation allele; genome mutation; trait phenotype; genotype Solution Albinism is a group of rare genetic disorders and termed as Oculocutaneous Albinism (OCA); wherein the affected person has little or no color in his skin, hair, or eyes. OCA is termed because the name of Oculocutaneous genes and classified as OCA; 1, 2, 3 and 4. OCA2 is also called the type II (tyrosinase-positive) OCA2. It is an autosomal recessive disorder and which is caused by reduced production of melanin pigment. The mutations related to OCA2 are three large deletions, six different point mutations and two small in-frame deletions. This mutation occurs in P gene. So, the fifth option is correct. Albinism is a phenotype because the disease is observable. And the mutation is associated to autosomal recessive disorder so, it is genotype. This will express when two recessive genes come together..

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You are presented with a patient, named Gerry, who has Down syndrome (trisomy 21). You determine that the nondisjunction event occurred in his mother, named Mary, during meiosis I. You are interested in the TPTE gene, which is near the centromere of chromosome 21. Virtually no recombination occurs between the centromere and the TPTE gene. Mary has two different alleles of the TPTE gene. Would Gerry\'s two chromosome 21s from his mother have the same or different alleles of the TPTE gene? Or do you not have enough information to decide? Explain your answer by drawing meiosis I and II for Mary\'s chromosome 21. Use \"A\" for one allele and \"a\" for the other allele of the TPTE gene. Only draw the chromosomes relevant to this question (Do not draw all 46 chromosomes). Solution a. Nondisjunction produces abnormal gametes. Nondisjunction in meiosis I occurs during anaphase I when one pair of homologous chromosomes fails to separate.Trisomy is the condition of having 3 copies of one chromosome type. It is designated as 2n + 1 because the cell has the normal two sets of each 23 types of chromosomes plus an extra copy of one chromosome. I think Gerrys two chromosome 21s from his mother have different alleles (A and a) of the TPTE gene. But more information could be needed to explain and understand this in a much better way..

You are presented with a patient, named Gerry, who has Down syndrome .pdf

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Write a template function to fill in a STL list with a given array. Solution Hi, I am not able to see the function prototype completely. But I wrote a function that takes a template array and adding all elements of array in STL list. Please let me know in case of any issue. template void fillArray(T arr[], int size) { // creating STL list of type T std::list myList; // adding each element from arr in list for(int i=0; i.

Write a template function to fill in a STL list with a given array..pdf

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write a method to generate AVL tree of height h wth fewest nodes and what is the running time of your method Solution class Node { int key, height; Node left, right; Node(int d) { key = d; height = 1; } } class AVLTree { static Node root; // A utility function to get height of the tree int height(Node N) { if (N == null) { return 0; } return N.height; } // A utility function to get maximum of two integers int max(int a, int b) { return (a > b) ? a : b; } // A utility function to right rotate subtree rooted with y // See the diagram given above. Node rightRotate(Node y) { Node x = y.left; Node T2 = x.right; // Perform rotation x.right = y; y.left = T2; // Update heights y.height = max(height(y.left), height(y.right)) + 1; x.height = max(height(x.left), height(x.right)) + 1; // Return new root return x; } // A utility function to left rotate subtree rooted with x // See the diagram given above. Node leftRotate(Node x) { Node y = x.right; Node T2 = y.left; // Perform rotation y.left = x; x.right = T2; // Update heights x.height = max(height(x.left), height(x.right)) + 1; y.height = max(height(y.left), height(y.right)) + 1; // Return new root return y; } // Get Balance factor of node N int getBalance(Node N) { if (N == null) { return 0; } return height(N.left) - height(N.right); } Node insert(Node node, int key) { /* 1. Perform the normal BST rotation */ if (node == null) { return (new Node(key)); } if (key < node.key) { node.left = insert(node.left, key); } else { node.right = insert(node.right, key); } /* 2. Update height of this ancestor node */ node.height = max(height(node.left), height(node.right)) + 1; /* 3. Get the balance factor of this ancestor node to check whether this node became unbalanced */ int balance = getBalance(node); // If this node becomes unbalanced, then there are 4 cases // Left Left Case if (balance > 1 && key < node.left.key) { return rightRotate(node); } // Right Right Case if (balance < -1 && key > node.right.key) { return leftRotate(node); } // Left Right Case if (balance > 1 && key > node.left.key) { node.left = leftRotate(node.left); return rightRotate(node); } // Right Left Case if (balance < -1 && key < node.right.key) { node.right = rightRotate(node.right); return leftRotate(node); } /* return the (unchanged) node pointer */ return node; } // A utility function to print preorder traversal of the tree. // The function also prints height of every node void preOrder(Node node) { if (node != null) { System.out.print(node.key + \" \"); preOrder(node.left); preOrder(node.right); } } public static void main(String[] args) { AVLTree tree = new AVLTree(); /* Constructing a tree with random values */ root = tree.insert(root, 10); root = tree.insert(root, 20); root = tree.insert(root, 30); root = tree.insert(root, 40); root = tree.insert(root, 50); root = tree.insert(root, 25); System.out.println(\"The preorder traversal of constructed tree is : \"); tree.preOrder(root); } } The running time of this method is O(h)..

write a method to generate AVL tree of height h wth fewest nodes and.pdf

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Which of the following is NOT correct? Which of the following is NOT correct?-Protozoans are fairly susceptible to antimicrobial controls, but their cysts are resistant.-Naked viruses are more susceptible to antimicrobial control agents than are enveloped viruses.-The antimicrobial effect of heat is enhanced by acidic conditions.-Warm disinfectants generally work better than cool ones. Solution The answer is -Naked viruses are more susceptible to antimicrobial control agents than are enveloped viruses..

Which of the following is NOT correctWhich of the following is NO.pdf

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What is Mill\'s response to the problem posed by some people\'s desire for virtue for its own sake?Tthis is except from Utilitarianism. Solution John Stuart Mill wrote his moral treatise, Utilitarianism, in a period in intellectual history dominated by Aristotelian virtue and Kantian deontology, as well as the intensely conservative political and social climate of Victorian England.Utilitarianism was intended to convince the public of Mill’s first principle. Mill responds to preexisting ethical frameworks, namely virtue ethics, deontology, Christian ethics, and egoism. Except for egoism, Mill invariably shows how other moral theories and values are actually or can become part of his first principle. The most prominent response in the text is his amalgamation of virtue ethics, which spans all of Chapter 4. Mill is successful in his synthesis of virtue ethics, despite philosophical sloppiness and logical fallacies. In his success, Mill crafted an impartial standard which the moral claims of other theories can by compared against. It is in Chapter 4 of Utilitarianism that Mill addresses virtue the most directly. He begins his argument by stating that virtue can actually be an end to human action besides happiness. Mill agrees with his opponents that there are people who desire or approve of acts according to virtue and absence of vice instead of the pleasure and the absence of pain. The reason this is significant is because if this is a descriptive fact, a truth, then it would pose a problem for Mill’s inductive proof of Utility, which hinges on the observation that happiness is desired and therefore desirable. If there are people who genuinely value virtue as the criteria of right and wrong, of how one should act, then, at very least, happiness is not the only first principle, which is what the Utilitarian claims. The solution to this problem for utilitarianism is to either swallow the intuitively unsavory conclusion that virtue is without value, or show that virtue actually has value belonging to, part of, or from happiness. Mill states that where opponents make their mistake is in believing that utility is incompatible with virtue. Mill holds that virtue is not only valuable as a means to pleasure, but in fact as an end in itself. Virtue and happiness are perceived to be different, but Mill attempts to show that this is superficial. It is by seeking virtue disinterestedly and for its own sake that virtue can become part of pleasure, Mill claims. So not only is Mill successful in bringing virtue ethics into utilitarianism, but he’s successful in creating a template by which other theories can be synthesized with utility. Mill’s claim is that all moral sentiments stem from a first principle that does not admit of proof, and that that first principle is the concept of utility. By providing one specific example, virtue, and showing how it actually admits of his general concept, utility, Mill is successful crafting an argument which allows people .

What is Mills response to the problem posed by some peoples desi.pdf

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What is a bivariate distributionSolutionBivariate distributio.pdf

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What are some of the tools you can use to tune or correct a Linux filesystem? Solution Some of the tools are Process Activity Command(it gives the real time view of system and we can keep an eye to factors which are affecting the performance.iptraf which gives the network statistics and other one could be /proc file system which gives the detailed information about harware devices connected and kernel information. Nagios is an open source system which give complete control over all the information like host, network traffic , equipments and it will send alert when something goes wrong. Another one could be Cacti which is web based monitoring tool, its a network graphing tool which gives the statistics in the form of graphs.

What are some of the tools you can use to tune or correct a Linux fi.pdf

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TrueFalse Insulin spikes in the postabsorptive state.Solution.pdf

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The principal role of ATP In the cell is: A) serving as structural support. B) catalyzing chemical reactions. C) keeping the phospholipid tails of membranes aligned D) providing energy. E) functioning as a hormone. Small nonpolar molecules, such as O_2 and CO_2, enter a cell by: A) vesicle-mediated transport. B) simple diffusion. C) osmosis. D) facilitated diffusion. E) active transport. Eukaryotic cells differ from prokaryotic cells in that eukaryotic cells have: A) plasma membrance B) genetic material C) cytoplasm. D) ribosomes. E) nucleus. If a plant cell is placed in an aqueous solution that is hypertonic to the cell, the cell A) build up turgor pressure B) build up wall pressure C) undergo plasmolysis D) neither gain nor lose water E) swell and perhaps burst Solution 15 . D Providing energy Atp or adenosine tri phosphate is a energy rich molecule produced by mitochondria of our cell. These atp molecules are produced by oxidation of organic molecules, (glucose) which is then utilised by our body for various cellular functions..

The principal role of ATP In the cell is A) serving as structural s.pdf

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The neutral theory of evolution proposed that most genetic mutations are functionally neutral. Neutral mutations are fixed in a population by natural selection random genetic drift environmental selection asexual reproduction the invariance of DNA Solution The neutra theory of evouloution proposed that most genetic mutations are functionally neutral.Neutral mutations are fixed in a population by- ans- E. the invariance of DNA (In a population genetics,fixation is the change in a gene pool with atleast two variant of particular gene or allele).

The neutral theory of evolution proposed that most genetic mutations .pdf

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The oldest living organisms on Earth are plants. Some bristlecone pines are over 4600 years old, and a desert creosote bush is known to be 10,000 year old. What special feature of plants provides for this incredible longevity? How do plant differ from animals in their pattern of growth and development? (also explain how trees grow upward as too how a nail driven into a 10year old tree will be at the same height 10 years later) Solution The oldest living organisms on Earth are plants. Some bristlecone pines are about 4,600 years old, and a desert creosote bush is known to be 10,000 years old. see actually it is just a matter of definition. no individual cells last that long time and all the original tissue is typically long dead. We generally think of it as continuing because the new growth is against the old. growth is continued and new cells are generated due to plants anaomy structure. they maintain their gene pool. trees grow upward as too how a nail driven into a 10year old tree will be at the same height 10 years later because tree grows from the top up. Differences between Plant Growth and Animal Growth: in plant growth , generally growth involved most of the parts and increase in the number of parts. growth generally continuw throughout the life of a plant. growth can be sessional. it depends on the seassion. it can quit the poisonius things during winter seasons. Plant growth carries well defined growing regions like stem, leaf. seedling has no effect in adult plant. growing pattern differs species to species in case of plant growth. new parts can be generated around the older ones. animal growth here growth takes place before the maturity period. after a certain age there is no growth in body organs. there is no increase in number of body parts. growth pattern is absent generally the young one is generally identical the old or adult one. ( sexual maturity may be an exception).

The oldest living organisms on Earth are plants. Some bristlecone pi.pdf

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Suppose there is a person whose total lung capacity (TLC) is 4.90 L, .pdf

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Research the IPv4 ns the IPv6 protocols, then prepare a report that explains the likes and differences of the two protocols and include how and why the transition from IPv4 to IPv6 is to occur. Project report body must have 5 pages or more, with Times New Roman12 pt. print double spaced. Also include a Cover sheet and a Reference page. Solution Introductory lines: The internet protocol or IP specifies and describes the technical details and formats of fragments of data (Also known as packets.) along with addressing schemes (numerical; actually in binary but can be interpreted as decimals or hexadecimals assigned to each of the devices connected through the network) in order to uniquely identify them and create a platform for secure and reliable communication. Ipv4 over the years was proved and deployed for communications over the internet. However in the current days it becomes obsolete and deficient in order to serve the huge and increasing number of users or network devices. To make this possible, the IPv6 comes into existence. The IPv6 provides many extra features and claims to support and serve this increasing number of devices for upcoming coming centuries. Let’s discuss about the similar and distinct features of both IPv4 and IPv6. IPv4 in terms of IP addresses, Address allocation, masks and types: IPv6 in terms of IP addresses, Address allocation, masks and types: IPv4 in terms of IP header: Headers in IPv4 are of variable length that varies from 20 to 60 bytes. IPv4 does not identify packet flow for QOS handling. It adds checksum filed for identification of transmission errors in the data. IPv6 in terms of IP header: Uses fixed length of 40 bytes of headers and there is no concept of option filed in header. It supports packet flow for QOS by flow level filed. IPv6 does not implements checksum for error handling. IPv4 in terms of IP packets ports and security: IPv4 supports the packet size of 576 bytes. They also support packet filtering mechanisms through firewalls. IPv4 supports for port spaces of the range 1 to 65535 IPsec in IPv4 is optional. IPv6 in terms of IP packets ports and security: IPv6 supports the packet size of 1280 bytes. They does not supports packet filtering mechanisms. The way the portts work both in IPv4 and IPv6 are the same. IPv6 has inbuilt IPsec support. IPv4 was successful;;y deployed in 1981 over internet whereas IPv6 was planned to be deployed from 1999 but could partially deployed till the date. Why transition from IPv4 to IPv6 must occur ? The deficiency of IP address of IPv4 type is one of the most eye catching reasons for transition towards IPv6. Each and every day the Internet is experiencing phenomenal increase of devices which are being connected and using the internet. In order to provide services to all the address space of IPv4 seems to be fewer. Hence a transition from IPv4 to IPv6 is must needed to satisfy such demand of larger address space. Except this, the other technical features like improve.

Research the IPv4 ns the IPv6 protocols, then prepare a report that .pdf

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Species I Species II Species III Species IV Ancestral sequence IV Sit.pdf

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Solve the differential equation y+y=sin(x). Im thinking I ne.pdf

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Please write a non-recursive procedure to print all nodes for a binary tree in a level order sequence. Solution #include #include struct node { int data; struct node* left, *right; }; void printGivenLevel(struct node* root, int level); int height(struct node* node); struct node* newNode(int data); void printLevelOrder(struct node* root) { int h = height(root); int i; for (i=1; i<=h; i++) printGivenLevel(root, i); } void printGivenLevel(struct node* root, int level) { if (root == NULL) return; if (level == 1) printf(\"%d \", root->data); else if (level > 1) { printGivenLevel(root->left, level-1); printGivenLevel(root->right, level-1); } } int height(struct node* node) { if (node==NULL) return 0; else { int lheight = height(node->left); int rheight = height(node->right); if (lheight > rheight) return(lheight+1); else return(rheight+1); } } struct node* newNode(int data) { struct node* node = (struct node*) malloc(sizeof(struct node)); node->data = data; node->left = NULL; node->right = NULL; return(node); } int main() { struct node *root = newNode(1); root->left = newNode(2); root->right = newNode(3); root->left->left = newNode(4); root->left->right = newNode(5); printf(\"Level Order traversal of binary tree is \ \"); printLevelOrder(root); return 0; }.

Please write a non-recursive procedure to print all nodes for a bina.pdf

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Please do C, D, and E For the given f: Z right arrow Z, decide whether f is onto and whether it is one-to-one. Prove that your decisions are correct. Solution c) onto as range of f(x) is Z and f(x) = odd intergers when x is even , f(x) = even intergers when x is odd Thus even integers + odd integers + 0 (when x is -1) = Z one to one as no element in x will give repeatable values as functions are increasing functions d) onto as x/2 covers whole range of intergers not one to one as as only x/2 covers the whole range (when x is even ) , repeated values will come when x is odd e) not onto as only multiples of 3 and 2 are getting in the range not one to one as it will be get repeated many times ( as in case 6 (for x=2) and 6 (for x=3)).

Please do C, D, and E For the given f Z right arrow Z, decide wheth.pdf

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need help with code I wrote. This code is a maze gui, and i need help with the method shorterstpath(); here is my code: ******mazeApp******* package mazepackage; import javafx.application.Application; import javafx.application.Platform; import javafx.scene.Group; import javafx.scene.Scene; import javafx.scene.canvas.Canvas; import javafx.scene.canvas.GraphicsContext; import javafx.scene.control.TextInputDialog; import javafx.scene.paint.Color; import javafx.stage.Stage; public class MazeApp extends Application { // default canvas size is DEFAULT_SIZE-by-DEFAULT_SIZE private static final int DEFAULT_SIZE = 768; private int width = DEFAULT_SIZE; private int height = DEFAULT_SIZE; // The graphics context is needed to enable drawing on the canvas private GraphicsContext gc; // boundary of drawing canvas, 0% border // private static final double BORDER = 0.05; private static final double BORDER = 0.00; private double xmin, ymin, xmax, ymax; public static void main(String[] args) { launch(args); } @Override public void start(Stage primaryStage) { Group root = new Group(); Canvas canvas = new Canvas(width, height); gc = canvas.getGraphicsContext2D(); gc.setLineWidth(2); gc.setFill(Color.WHITE); gc.fillRect(0, 0, width, height); root.getChildren().add(canvas); TextInputDialog tid = new TextInputDialog(); tid.setTitle(\"Maze Size\"); tid.setHeaderText(\"Enter maze size between 10 and 50\"); tid.showAndWait(); int size = Integer.parseInt(tid.getResult()); if (size > 50) size = 50; if (size < 10) size = 10; primaryStage.setTitle(\"Maze Application\"); primaryStage.setScene(new Scene(root)); primaryStage.setResizable(false); // Make sure that the application goes away when then window is closed primaryStage.setOnCloseRequest(e -> System.exit(0)); primaryStage.show(); Maze maze = new Maze(this, size); // Must solve the maze in a separate thread or else // the GUI wont update until the end..... Thread solver = new Thread( new Runnable () { public void run() { while(true) { maze.buildAndDrawMaze(); maze.findShortestPath(); try { Thread.sleep(5000); } catch(Exception e) { } gc.setFill(Color.WHITE); gc.fillRect(0, 0, width, height); } } }); solver.start(); } /** * Sets the pen color to the specified color. * * @param color the color to make the pen */ public void setPenColor(Color color) { gc.setStroke(color); } /** * Sets the pen color to the specified color. * * @param color the color to make the pen */ public void setFillColor(Color color) { gc.setFill(color); } /** * Sets the x-scale to the specified range. * * @param min the minimum value of the x-scale * @param max the maximum value of the x-scale * @throws IllegalArgumentException if {@code (max == min)} */ public void setXscale(double min, double max) { double size = max - min; if (size == 0.0) { throw new IllegalArgumentException(\"the min and max are the same\"); } xmin = min - BORDER * size; xmax = max + BORDER * size; } /** * Sets the y-scale to the specified range. * * @param min the minimum valu.

need help with code I wrote. This code is a maze gui, and i need hel.pdf

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Microbiology question : Which of the following is the most unique aspect of archaeon cell envelopes? a) The plasma membrane is composed of phospholipids. b) The cell walls contain a modified type of peptidoglycan. c) Archaeon capsules can be composed of protein or polysaccharide. d) Some archaeons have lipid monolayers in their plasma membrane. Solution The distinct features is Some archeons have a lipid monolayer in their plasma memberane..!!!.

Microbiology question Which of the following is the most unique a.pdf

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Is Event B dependent or independent of Event A A green ball is draw.pdf

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