Viruses might be an example of what is called an evolutionary stable .pdf

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Viruses might be an example of what is called an evolutionary stable strategy. Describe the adaptive advantages of the viral \"way of life\" and how selection contributes to this stability. How is a repressible operon different than an inducible operon? Solution 1. Viruses are acellular, microscopic and obligate parasites that require a host for their survival. Viruses can infect plants, animals and bacteria too. The viruses that infect bacteria are called bacteriophages. The diagram shows the bacteriophage (T7), which is in tadpole shape. The tobacco mosaic virus (TMV) shown in the diagram infects tobacco plants is a rod shaped virus. The human immune deficiency virus (HIV) is in spherical shape. Most of the plant viruses contain RNA as genetic material with few exception of having DNA. The bacteriophages contain DNA with few exceptions. The viruses that infect animals and humans generally have DNA but the HIV contains RNA.Though, the viruses have different forms and shapes, all viruses contain a nucleic acid either DNA or RNA, which is surrounded by a protein capsid or coat or shell. Some of the viruses have surface glycoproteins that act as antigens and elicit immune response in host. The immune system of the host try to eliminate them by producing antibiotics or treatment with antibiotics but the viruses can change their antigentic structures through mutations, genetic shift and drift and are able to cause infection in same host. Thus, viruses have new strategies always for their evolution and have adoptations and advantages to win over host to maintain a stable evolutioary strategy. 2. Genes in prokaryotes are regulated in clusters called operons. An operon contains a set of genes consisting of promoter, operator, and a cluster of structural genes that produce products having same function. The operon can be upregulated or down regulated to control its expression. Repressor is a protein that inhibits the transcription of genes when it is bound to specific DNA sequences or near promoter. The repressor is bound at the operator to inhibit the transcription. For example, in tryptophan operon when tryptophan is available in excess amounts, the repressor bind to operator to turn off the transcription. In inducible operon, a small molecule called inducer binds to operator to turn on the transcription of the genes. In absence of inducer, repressor binds to operatyor and inhibits transcription. In Lac operon, lactose acts as inducer that brings about confirmational change in promoter and thereby RNA polymerase binds to it to initiate transcription..

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Studying Viruses can be important because they are often disease causing pathogens. Howeve, because of how easy they are to control and mainpulate in the laboratory, viruses can be good tools to study basic, fundamental principles of cell and molecular biology. Concisely explain one example of how viruses were used to make a critical molecular genetics discovery (briefly, how was the experiment done, and what was concluded?) Studying Viruses can be important because they are often disease causing pathogens. Howeve, because of how easy they are to control and mainpulate in the laboratory, viruses can be good tools to study basic, fundamental principles of cell and molecular biology. Concisely explain one example of how viruses were used to make a critical molecular genetics discovery (briefly, how was the experiment done, and what was concluded?) Solution Viruses are intracellular parasites that are effective after infecting particular host cells. Viral infection starts offevolved when proteins on the surface of a virion bind to particular receptor proteins at the host cells. The specificity of this interaction determines the host range of an endemic. Aside from being the causative agents of many sicknesses, viruses are vital gear in cellular biology research, mainly in research on macromolecular synthesis. Individual viral particles (virions) usually incorporate both an RNA or a DNA genome, surrounded via multiple copies of 1 or a small variety of coat proteins, forming the nucleocapsid. The nucleocapsid of many animal viruses is surrounded with the aid of a phospholipid bilayer, or envelope. During lytic replication, host-cell ribosomes and enzymes are used to explicit viral proteins, which then viral genome and package it into viral coats. The multiple progeny virions produced within a single infected cellular eventually are launched, following cellular lysis or gradual disintegration of the cell . Progeny nucleocapsids of enveloped viruses are launched by using budding of the host-cell membrane in which viral membrane proteins have been deposited. Some bacterial viruses (bacteriophages) might also go through lysogeny following infection of host cells. In this situation, the viral genome is incorporated into host-cell chromosomes, forming a prophage this is replicated along with the host genome. When suitably activated, a prophage enters the lytic cycle. All retroviruses and some different animal viruses can combine their genomes into host-cell chromosomes . In a few instances, this leads to abnormal cell replication and the eventual improvement of cancers. Recombinant viruses can be used as vectors to transduce genes into cells. In this method, viral genes required for the lytic cycle are changed with the aid of other genes. The use of viral vectors for gene remedy continues to be in its infancy, but has super ability for treatment of diverse diseases..

match the following attributes to the parts of a compilerstrips ou.pdf

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match the following attributes to the parts of a compiler strips out the comments and whitespace converts text into lexemes generates an Abstrat Symbol Tree recursive descent or table driven uses BNF or EBNF definitions a. lexer b. parser Solution The Program and Portfolio Management Maturity Model is an effective tool for organizations to decide quickly what PPM improvements they should make to enhance their organization\'s ability to optimize investments, execute big changes and deliver value. Overview Key Findings Real project, program and portfolio management, when operating at the level of well-integrated practices, is the key enabler that allows organizations to identify and execute strategic change. Any meaningful undertakings to enhance or evolve the program and portfolio management (PPM) function must pay more than lip service to organizational structure, business model and culture to have any chance of success. Competitive pressures and changing market conditions are forcing organizations toward Level 3, and most organizations still aren\'t ready to make the leap. Recommendations At any level of PPM maturity, focus first on helping the organization make well-informed investment choices. This will improve the odds of project success more than any other factor. For organizations that require enterprisewide change and capabilities, it is worth the effort to pursue Level 4, where enterprise PPM practices are built. Identify objectives for your enterprise, and use them to identify the most salient improvement opportunities for your enterprise. Meticulously manage the change involved in maturing/improving your PPM capabilities. Analysis Project/program/portfolio management office (PMO) leaders, program managers and portfolio managers are frequently challenged by ineffective processes, lack of stakeholder engagement and difficult-to-quantify value. 1 Part of the problem is failure to match their processes, people and technology approaches to the maturity level of their organizations. 2 The Gartner Program and Portfolio Management (PPM) Maturity Model assessment is designed to help PPM leaders understand best practices around large project management, as well as PPM to handle delivery of strategic value. This model assumes that organizations progress through a maturity curve and that each level of organizational maturity directly affects the level of investment and types of PPM approaches organizations choose to adopt. arsing A parser is an algorithm that determines whether a given input string is in a language and, as a side-effect, usually produces a parse tree for the input. There is a procedure for generating a parser from a given context-free grammar. Recursive-Descent Parsing Recursive-descent parsing is one of the simplest parsing techniques that is used in practice. Recursive-descent parsers are also called top-down parsers, since they construct the parse tree top down (rather than bottom up). The basic idea of recursive-descent par.

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match the following attributes to the parts of a compilerstrips ou.pdf

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Looking for some help with this assignment. 1. Define distributed computing and the impact it has had on organizations. 2. Why should you be concerned about the innovation butterfly is you are an IT professional? 3. What is the difference between single threading and multithreading and how would I use it in a Distributed system? 4. What is a mobile architecture and what are some of the design considerations? 5. Explain why naming conventions are important in large enterprise organizations. 6. What is the difference between a disruptive and a sustaining technology? 7. Describe the difference amongst Mobile, Cloud and Embedded systems. 8. Explain the role time plays in distributed systems and why it is difficult to manage. 9. What are the major challenges IOT plays in a distributed system? 10. How does mobility, ubiquity and PAN increase the complexity of distributed systems. Solution 1) sol:defnition: it is a one type of model where software components are shared with mutliple computer to make work easier and gives more effieciency and performance impact of distributed computing 1)at present it keeps a main roles in organizations ,all the organizations are converting to distributed system 2)we can process data at one place and access data at one place so there is lot of availability 3)many organization doing thier business process in remote computer and accesing their database 2)differenece between sustaining and disruptive technology 1)by using sustaining technology improves performance for products which had been established with the dimension of the product which gives market to the mainstream customers 2) coming to the disruptive technology bring to market in diffrent praportions of values which had been available previously,these disruptive techonolgy underperform which had been established in main steram market.

Introduction and BackgroundIn recent lectures we discussed usi.pdf

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Introduction and Background In recent lectures we discussed using arrays, classes and interfaces (see newly added course notes if you want to read ahead about interfaces – we will cover them this week in lecture). In this lab you will utilize all of these topics to build a simple yet useful new class. Consider the following interface describing the methods for a simple double ended queue (or deque): public interface SimpleDeque { public void addFront(Object X); // Add Object X at front of list public void addRear(Object X); // Add Object X at rear of list // If array is full, add methods should do nothing public Object removeFront(); // Remove and return Object X from // front of list public Object removeRear(); // Remove and return Object X from // rear of list // If array is empty, remove methods should return null public boolean isEmpty(); // Return true if the list is empty // Return false otherwise } A queue has the behavior such that items are added at the rear and removed from the front, thereby giving a First In First Out (FIFO) access to the items added and subsequently removed from the list. No other manipulations of the data are permitted (for example, we cannot add or remove anywhere in the middle). Looking at it \"in reverse\", we could add new items at the front of the queue and remove them from the rear. This is still providing FIFO access, but just from a different point of view. Now consider both adding and removing items at the rear of the list (without ever accessing the front). This is called stack access and gives us Last In First Out (LIFO) access to the items (the data is removed in reverse order). The same behavior occurs if we both add and remove at the front without ever accessing the rear of the list. The simple deque above is expressed as an interface rather than a class, because we are not describing the data or how it is represented -- we are simply describing its access behavior. However, to actually build a working deque, we need a class that implements the interface above. For example: public class MyDeque implements SimpleDeque { Object [] theData; int numItems; public MyDeque(int maxItems) { theData = new Object[maxItems]; numItems = 0; } // Implementation of the five methods of SimpleDeque, plus // perhaps other methods as well } Note that the implementation above uses an array of Object to store the items in the deque. Since Object is the base class to all other Java classes, an array of Object can thus be used to store any Java class types (we can even store primitive values if we utilize their wrapper classes). Also note that nothing in the SimpleDeque interface requires an array to be used to store the data. You will see in your CS 0445 course that a linked list may in fact be a better implementation than an array in this case. However, for this implementation we will use an array because it is simple and easy to understand. Another important thing to notice about the partial implementation above is tha.

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In the binary search, if the array being searched has 32 elements in it, how many elements of the array must be examined to be certain that the array does not contain the key? What about 1024 elements? Note: the answer is the same regardless of whether the algorithm is recursive or iterative. Solution Binary Search Algorithm- Fundamentals, Implementation and Analysis Hitesh Garg | May 15, 2015 | algorithms | 5 Comments Binary Search Algorithm and its Implementation In our previous tutorial we discussed about Linear search algorithm which is the most basic algorithm of searching which has some disadvantages in terms of time complexity,so to overcome them to a level an algorithm based on dichotomic (i.e. selection between two distinct alternatives) divide and conquer technique is used i.e. Binarysearch algorithm and it is used to find an element in a sorted array (yes, it is a prerequisite for this algorithm and a limitation too). In this algorithm we use the sorted array so as to reduce the time complexity to O(log n). In this, size of the elements reduce to half after each iteration and this is achieved by comparing the middle element with the key and if they are unequal then we choose the first or second half, whichever is expected to hold the key (if available) based on the comparison i.e. if array is sorted in an increasing manner and the key is smaller than middle element than definitely if key exists, it will be in the first half, we chose it and repeat same operation again and again until key is found or no more elements are left in the array. Recursive Pseudocode: 1 2 3 4 5 6 7 8 9 10 11 12 // initially called with low = 0, high = N – 1 BinarySearch_Right(A[0..N-1], value, low, high) { // invariants: value >= A[i] for all i < low value < A[i] for all i > high if (high < low) return low mid = low +((high – low) / 2) // THIS IS AN IMPORTANT STEP TO AVOID BUGS if (A[mid] > value) return BinarySearch_Right(A, value, low, mid-1) else return BinarySearch_Right(A, value, mid+1, high) } Iterative Pseudocode: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 BinarySearch_Right(A[0..N-1], value) { low = 0 high = N - 1 while (low <= high) { // invariants: value >= A[i] for all i < low value < A[i] for all i > high mid = low +((high – low) / 2) // THIS IS AN IMPORTANT STEP TO AVOID BUGS if (A[mid] > value) high = mid - 1 else low = mid + 1 } return low } Asymptotic Analysis Since this algorithm halves the no of elements to be checked after every iteration it will take logarithmic time to find any element i.e. O(log n) (where n is number of elements in the list) and its expected cost is also proportional to log n provided that searching and comparing cost of all the elements is same Data structure used -> Array Worst case performance -> O(log n) Best case performance -> O(1) Average case performance -> O(log n) Worst case space complexity -> O(1) So the idea is- RECURSIVE Implementation of Binary search in C programming language 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 .

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Heliconius melpomene is a species of butterfly with ZW sex determination. H. melpomene\'s wing spots are normally orange, but a recessive Z-linked mutation causes the spots to be yellow, while a recessive autosomal mutation causes the spots to be red. Butterflies homozygous (or hemizygous) for both mutations have no wing spots. What are the expected F1 and F2 phenotypes from a cross between a red-spotted female and a yellow-spotted male? Solution The Heliconius melpomene butterfly having sex determination ZW have normally orange wing spots. Given Recessive Z linked mutation causes the spots to be yellow, if Y is the dominant gene the y represents the recessive gene. The recessive autosomal mutation causes the spot to be red,the recessive autosomal mutation represented by rr. a) WX represents female b) WW represents male Given-: F1 generation The red spotted female is crossed with yellow spotted male ZWrr X ZZy ZZ, WrrZ, ZZy, WrrZy It gives 1 hemizygous butterfly, 1red spotted female butterfly, 1 yellow spotted male and 1 normal orange wing male butterfly. The F2 generation gives the same result..

Identify the key mechanism that determines which X remains active an.pdf

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Given to girls born in the U.S. in the years The tables below give the ten most popular names 1974 and 1999 and the percentages of girls given those names. Consider group of 20 girls born in 1974. and another group of 20 bom in 1999. Use the data in the tables to do problems 22-25 regarding these groups. The name Jennifer was approximately 3 times as popular in 1974 as Emily was in 1999, despite each being the #1 name in their respective years. Determine by calculation if you are 3 times as likely to have more than one Jennifer in the 1974 group of 20 as you are to have multiple Emilys in the 1999 group of 20. Solution.

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Ferdinand is affected by hemophilia, a recessive x-linked condition..pdf

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Describe the authentication process in the figure below Solutio.pdf

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