In order to determine the overall chromosome map for E. coli, four Hf.pdf

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In order to determine the overall chromosome map for E. coli, four Hfr strains (#1, 2, 3, & 4) were obtained from F+ ancestors. The start times of transfer during conjugation matings with F- strains carrying auxotrophic and fermentation markers were determined for each Hfr strain, as listed. Using the times for transfer of each maker, draw a unified chromosome map of E. coli, showing the order and relative distances between these markers. Place the origin of transfer (site of F insertion) on the map for each Hfr strain. What is the total length of the E. coli chromosome, in transfer minutes? Solution The process of assigning genes to particular locations on a chromosome is known as chromosome mapping and the map is called gene map. There are many methods used by scientists to find out the correct location by mapping genes. By using these methods different chromosome maps can be prepared like cytogenetic maps, linkage maps, DNA sequence maps or physical maps. Hfr strain indicates a high frequency of recombinants. The polarity of the Hfr chromosome is determined by the orientation in which F+ is inserted in it. The relative position of chromosomes on a gene can be found out by linkage maps which are measured in time taken by second gene in following the first one during conjugation. Here it is assumed that genes separated by 10 minutes near the entry end are physically same distance apart as the genes separated by 10 minutes at the F attachment site..

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A microbial geneticist isolates a new mutation in E. coli and wishes to map its chromosomal location. She uses interrupted-mating experiments with Hfr strains and generalized-transduction experiments with phage P1. Explain why each technique, by itself, is insufficient for accurate mapping. Solution Hfr cells undergoing conjugation transfer host genes in a linear fashion. The genes transferred depend on both the Hfr strain and the length of time during which the transfer occurred. F factors arise from improper excision of an Hfr from bacterial chromosome. They can have only specific bacterial genes on them because the integration site is fixed for each strain. P1 does not recognize pseudo-pac sites on the chromosome. P1 seems to begin packaging from random double-stranded breaks on the chromosome that are generated during phage infection. Generalized transduction is the process by which any bacterial gene may be transferred to another bacterium via a bacteriophage, and very rarely a small number of phages carry the donor. A restricted set of genes will not be able to transfer by this method. She wanted to transfer a specific gene to a specific located which is not doen by generalized transduction, so accurate gene mapping will be difficult by these methods..

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Consider a binary SNP matrix M with 1 representing the derived allele, and 0 as the ancestral allele. The SNP matrix has n haplotypes. The HAF vector of a haplotype h in M, denoted c_h, is obtained by taking the binary haplotype vector and replacing non-zero entries (derived alleles carried by the haplotype) with their respective frequencies in the sample. Define the HAF score of h as: HAF (h) sigma_j c_h[j] where the sum proceeds over all segregating sites j in the genomic region. Derive an expression for the Expected value of HAF (h) for h randomly sampled from a neutrally evolving sample of n individuals. Solution SNP (Single Nucleotide Polymorphism) Frequency of SNPs is greater than that of any other type of polymorphism. Having the understanding of the structure and frequencies of haplotypes is very important for associating genetic polymorphisms with a given trait. It also helps in inferring the genetic genealogy of alleles in a population. Single nucleotide polymorphism (SNP) haplotypes can be determined without ambiguity when an individual does not have more than one heterozygous site in a given genomic region. The Haplotype Allele Frequency (HAF) score assigned to individual haplotypes in a sample naturally captures many of the properties shared by haplotypes carrying a favored allele. Haplotype: Combinations of genetic variants occurring on the same DNA molecule. Each gene in the diploid genome has two sequences, one on each haplotype. Methods for detecting the genomic signatures of natural selection have been heavily studied. They have been successful in identifying many selective sweeps. For most of these sweeps, the favored allele remains unknown, making it difficult to distinguish carriers of the sweep from non-carriers. In an ongoing selective sweep, carriers of the favored allele are likely to contain a future most recent common ancestor. a) Expected value of HAF (h) for h randomly sampled from a neutrally evolving sample of n individuals. Consider a neutrally evolving sample of n individuals. We assume that all sites are biallelic, and at each site, we denote ancestral alleles by 0 and derived alleles by 1. We also assume that all sites are polymorphic in the sample. The HAF vector of a haplotype h, denoted by c, is obtained by taking the binary haplotype vector and replacing non-zero entries (derived alleles carried by the haplotype) with their respective frequencies in the sample. For parameter , we define the - HAF score of c as: where the sum proceeds over all segregating sites j in the genomic region. The 1-HAF score of a haplotype amounts to the sum of frequencies of all derived alleles carried by the haplotype. The - HAF score is equivalent to the -norm of c raised to the th power..

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Hello, I need some assistance in writing a java program THAT MUST USE STACKS (LIFO). Create a Calculator w/ GUI Write a program that graphically displays a working calculator for simple infix expressions that consist of: single-digit operands, the operators: +, -, *, and /, and parentheses. Make the following assumptions: unary operators (e.g. -2) are illegal all operations, including division, are integer operations (and results are integers) the input expression contains no embedded spaces and no illegal characters the input expression is a syntactically correct infix expression division by zero will not occur (consider how you can remove this restriction) Create a GUI application, the calculator has a display and a keypad of 20 keys, which are arranged as follows: C < Q / 7 8 9 * 4 5 6 - 1 2 3 + 0 ( ) = As the user presses keys to enter an infix expression, the corresponding characters appear in the display. The C (Clear) key erases all input entered so far; the < (Backspace) key erases the last character entered. When the user presses the = key, the expression is evaluated and the result appended to the right end of the expression in the display window. The user can then press C and enter another expression. If the user presses the Q (Quit) key, the calculator ceases operation and is erased from the screen. C < Q / 7 8 9 * 4 5 6 - 1 2 3 + 0 ( ) = Solution import java.awt.*; import javax.swing.*; import java.awt.event.*; public class Calculator extends JFrame implements ActionListener { JPanel[] row = new JPanel[5]; JButton[] button = new JButton[19]; String[] buttonString = {\"7\", \"8\", \"9\", \"+\", \"4\", \"5\", \"6\", \"-\", \"1\", \"2\", \"3\", \"*\", \".\", \"/\", \"C\", \"\", \"+/-\", \"=\", \"0\"}; int[] dimW = {300,45,100,90}; int[] dimH = {35, 40}; Dimension displayDimension = new Dimension(dimW[0], dimH[0]); Dimension regularDimension = new Dimension(dimW[1], dimH[1]); Dimension rColumnDimension = new Dimension(dimW[2], dimH[1]); Dimension zeroButDimension = new Dimension(dimW[3], dimH[1]); boolean[] function = new boolean[4]; double[] temporary = {0, 0}; JTextArea display = new JTextArea(1,20); Font font = new Font(\"Times new Roman\", Font.BOLD, 14); Calculator() { super(\"Calculator\"); setDesign(); setSize(380, 250); setResizable(false); setDefaultCloseOperation(EXIT_ON_CLOSE); GridLayout grid = new GridLayout(5,5); setLayout(grid); for(int i = 0; i < 4; i++) function[i] = false; FlowLayout f1 = new FlowLayout(FlowLayout.CENTER); FlowLayout f2 = new FlowLayout(FlowLayout.CENTER,1,1); for(int i = 0; i < 5; i++) row[i] = new JPanel(); row[0].setLayout(f1); for(int i = 1; i < 5; i++) row[i].setLayout(f2); for(int i = 0; i < 19; i++) { button[i] = new JButton(); button[i].setText(buttonString[i]); button[i].setFont(font); button[i].addActionListener(this); } display.setFont(font); display.setEditable(false); display.setComponentOrientation(ComponentOrientation.RIGHT_TO_LEFT); display.setPreferredSize(displayDimension); for(int.

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Explain how consumer judges the quality of service (Including price as indicator of quality) Solution It is more difficult for consumers to evaluate the quality of service than the quality of products. This is true because of certain distinctive characteristics of services; They are intangible, they are variable, they are perishable, and they are simultaneously produced and consumed. To overcome the fact that consumers are unable to compare competining services side-by-side as they do with competing pproducts, consumers rely on surrogate cues (i.e., extrinsic cues) to evaluate the service quality. It must be noted here that the service quality is a perception of the consumer and not described in absoolute terms like in the case of products. Perceived service value has been described as a trade-off between the perceived benefits (quality) of service and the perceived sacrifice - both monetary and non-monetary - necessary to acquire it. Consumers rely on price as an indicator of service quality. Consumers using a price / quality relationship are actually relying on a well-known (and, hence, more expensive) brand name as an indicator of quality without actually relying directly on price per se. Consumers also use price and brand to evaluate the prestige of the product but do not generally use these cues when they evaluate the service delivery and performance. Since, price is so often considered as an indicator of quality, consumers perceive a service to be of higher quality based on the pricing of the service. However, on other hand, when the consumers evaluate the service more critically, then they rely less on the price and brand as indicators of service quality..

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Early Americans did not practice equality according to today’s definitions; however, the concept of equality has always been an important part of American rhetoric with an evolving definition. Politicians considered women, blacks, and the lower classes as “dependent”, as they owed their livelihood to someone else. They were thought not to have the “independence” to form conclusions. How is Mercy Otis Warren an example against this theory? Solution Mercy Otis Warren was a political writer and propagandist of the American Revolution. In the eighteenth century, topics such as politics and war were thought to be the province of men. Few men and fewer women had the education or training to write about these subjects. Warren was an exception. During the years before the American Revolution, Warren published poems and plays that attacked royal authority in Massachusetts and urged colonists to resist British infringements on colonial rights and liberties. During the debate over the United States Constitution in 1788, she issued a pamphlet, Observations on the new Constitution, and on the Federal and State Conventions written under the pseudonym \"A Columbian Patriot,\" that opposed ratification of the document and advocated the inclusion of a Bill of Rights. Observations was long thought to be the work of other writers, most notably Elbridge Gerry. It was not until her descendant, Charles Warren, found a reference to it in a 1787 letter to British historian, Catharine Macaulay, that Warren was accredited authorship.In 1790, she published a collection of poems and plays under her own name, a highly unusual occurrence for a woman at the time. In 1805, she published one of the earliest histories of the American Revolution, a three-volumeHistory of the Rise, Progress, and Termination of the American Revolution, the first history of the American Revolution authored by a woman..

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Design a circuit that detect the sequence 0110. Draw the Moore state machine for detecting non-overlapping sequences Write the truth table that shows the next state based on T flip flops Draw the circuit Re-do the More state machine graph for overlapping sequences (do the state machine only) Solution A)This sort of situation might arise for a simple code lock, where the user must enter the correct 4 bits to open the lock. If the user doesn’t input the correct 4 bits, they must start over. Or, it might be used in a simple communication system which receives bits off a line one at a time, and the word size is 4 bits. In this case, the patterns might be start and stop signals, or some other communication protocol-related information. The input is defined as X. Thus, the value of X is the value of the input line in a given clock cycle. The output, Y, goes high for 1 clock cycle as soon as it receives the 4th bit that matches a pattern. We have only one input and only one output, so we don’t need to use ASM charts. We can use a simple state diagram. Note that because the output must go high as soon as the 4th matching bit is received, we need this to be a Mealy machine. Also, note that in this example, when we are looking for 1010, we assume the most significant bit is the first bit received, so the order of the inputs would be 1-0-1-0, not 0-1- 0-1. Our state machine starts in a state in which we have received no bits. We will call this state START. The state transitions will be depending on whether the input X is a 0 or 1 – we can essentially use the states to record which values have been received. If the input is 0, we transition to a state called S0; if X is 1, we transition to S1. These states are named after the value received in the previous clock cycle. From each of S0 and S1, we can likewise go to new states based on the value of X. If in S0, and X is 0, we go to state S00; if X is 1, we go to state S01. Similarly, if in S1, and X is 0, we go to state S10; if X is 1, we go to state S11. Again, the states are named after the two previously received values. Now, each of those four states can again go to one of two possible states, based on the value of X. The states are S000, S001, S010, S011, S100, S101, S110, S111. At this point, the state machine knows we have received 3 bits, and knows what those 3 bits are. When it receives the next bit, it can determine whether a matching sequence was input, or whether we need to restart. So, S000, S001, S010, S100, S110, and S111 all transition back to START and output 0, no matter what the input, because none of those states correspond to having received the first 3 bits of sequences 0110 or 1010. States S011 and S101, however, do depend on the input. They both still transition back to START, but if the input is 0, the output is 1. Otherwise, the output remains 0. Thus the state machine meets our design requirements. So we have a state machine that has 15 states. 15 states is quite a lot. This requ.

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Which of the following characteristics of the fetus distinguishes it from the embryo? Select one: a. Blood cells form and fill primitive blood vessels. b. The primitive streak appears. c. Bone replaces the softer cartilage. d. The specialization of one group of cells causes adjacent groups of cells to specialize. Solution D. The specialization of one group of cells causes adjacent groups of cells to specialize. this is called differentiation of cells.

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