2. A one-period rate of return from an investment can be measured as.pdf

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2. A one-period rate of return from an investment can be measured as: Rate of Return (Ending Value-Beginning Value)+ Income/Beginning Value. If a share of stock goes from $20 to $22 in one year and also pays a dollar in dividends during the year, what is the rate of return for the period? (L01-1) Solution Rate of return = ( Ending value + dividends - Beginning value) / beginning value Rate of return = ( 22 + 1 - 20 ) / 20 Rate of return = 3 / 20 Rate of return = 0.15 or 15%.

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A mutant strain of E. coli is found that produces both beta-galactosidase and permease at high levels whether lactose is present or not. (a) What are two possible genotypes for this mutant? Another mutant is isolated that produces no beta-galactosidase at any time but produces permease if lactose is present in the medium. A merodiploid strain is formed from these two mutants. in the absence of lactose, neither beta-galactosidase nor permease is made. When lactose is added, the merodiploid strain makes both beta-galactosidase and permease. (b) What is the genotype of the merodiploid strain? Solution A mutant of E. coli is found that produces both beta - galactosidase and permease at high levels in the presence or absence of lactose a) The two possible genotypes for this mutant is given as-: F+ Z+ Y+ Now another mutant is isolated that produces no beta-galactosidase at any time but produces permease if lactose is present in the medium. The genotype for this mutant is given as-: F- Z- Y- Given a merodiploid strain is formed from these two mutants. Merodiploid is partially diploid bacterium having its own chromosome and a fragment that is introduced either by conjugation, transformation or transduction. It is given by-: F+ Z+ Y+ x F- Z- Y- =F\' Z+ - Y- b) The genotype for the merodiploid strain is F\' Z+ - Y- For Merodiploid strain in the absence of lactose, neither beta-galactosidase nor permease is formed but in the presence of lactose both beta-galactosidase and permease is produce. As mutant 1 have the ability to make beta-galactosidase in the absence or presence of lactose but the mutant 2 is unable to make beta-galactosidase at any time so in the new mutant the gene for beta galactosidase is expressed only in the presence of lactose. Also the permease producing gene of 2 mutant is dominant that is the reason the merodiploid strain produces permease in the presence of lactose but not in the absence of lactose..

A mutant strain of E. coli is found that produces both beta-galactos.pdf

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An unknown individual launches a series of attacks against the Web sites of Prime Sales Corporation. The attacks significantly slow the sites, leading to $100 million in damage in terms of lost work time, lost revenue, site repair costs, and other expenses. The attacker does not intend to profit from the onslaught and in fact does not realize any financial benefit from the effects. How is this attack most likely orchestrated? Who is most likely to engage in this act—that is, whose habits and limitations are clearly suited to such conduct—and why? Solution Most probably the website is slowed by a DDOS (Distributed Denial of Service) attack. In this attack the hacker sends multiple fake requests from different servers with the intention of slowing down the service to legitimate users. The most likely suspect is an anonymous hacking group. These groups are formed by professional hackers who come together for a common cause. The most likely reason might to gain attention or the group considers something the company is doing as against their beliefs..

An unknown individual launches a series of attacks against the Web s.pdf

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You\'ve completed an interrupted mating experiment in E. coli, the leu gene enters before the aln gene. For this experiment, you crossed an Hfr that is leu^+ gln^+ with an F- that is leu^- gln^-. Exconjugant cultures are plated on minimal medium supplemented with leonine but not glutamine, and 250 total colonies were observed. Each of those colonies was then plated on minimal medium containing no leonine or glutamine; only 200 of the cultures grew under these conditions. What are the genotypes of the two different cultures? So, what likely happened at the cellular/molecular level to allow some cells to grow without glutamine supplementation, but not all? Make sure you explain/support your answer. (lessthanorequalto 4 sentences) Solution a) leu and gln gene\'s are the genotypes of the two different cultures. b) because of this minimal medium supplemented and Hfr , F for combination of any two moleucles or ions the opposite signs of its bonding makes stable for a molecule. so, this molecules will allow.this makes cellular/moleculer level to allow some cells to grow without glutamine supplementation..

Youve completed an interrupted mating experiment in E. coli, the l.pdf

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Write the following Scheme functions in Java: Solution import java.util.ArrayList; import java.util.Arrays; import java.util.List; public class Neshted { public static String neshted_list(List list){ String output = \"\"; float count =0; for(int i : list){ if(count == list.size()/2){ if(list.size() %2 == 0) output += \"()\"; else output += \"(\"; } if(count=list.size()/2) { output += i+\")\"; } count ++; } return output; } public static void main (String args[]){ Integer[] values = {4,5,3,2,8}; List list = Arrays.asList(values); System.out.println(neshted_list(list)); } }.

Write the following Scheme functions in JavaSolutionimport ja.pdf

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Write a C program to enter a first name, middle name and last name in separate variables. After entering each name at the keyboard, call a function named convert that will capitalize the first letter of the name and set the rest of the name to lowercase (for example, jOhN becomesJohn). The convert function will be called three times-- once for the first name, once for the middle name, and once for the last name. Combine the properly capitalized names into one string and output the string (make sure you have a space between each part of the name). Count the number of characters in the total name (including spaces between names) and output that total also. Include descriptive comments in your source code. Solution #include #include #include void Convert(char string[]) { int i; int len = strlen(string); //compute length of string string[0] = toupper(string[0]); // capitalize first character of the string for (i=1;i.

Write a C program to enter a first name, middle name and last name i.pdf

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Why do the soft technologies open more opportunities for women? To what extent have these technologies impacted the perceptions of men’s and women’s roles in the economy, within marriage, and in society as a whole? Solution Now a days Woman is playing a very important role in every field of the economy.Either Agriculture or education or any part of the manufacturing area. I am also agree from this statement, beacuse after marriage so many responsibilities increase its libilities through these soft techniqe she can hendle her all task very easily. Technology is a vital element for any community. It plays a significant role in food security, agriculture and small scale production. Women use technical skills and knowledge in their daily activities; they continually innovate and adapt technologies in response to things they face in their everyday lives. However, their role and technical skills are often overlooked and undervalued. There are four main reasons why women are less visible in the application of technology than men:.

Why do the soft technologies open more opportunities for women To w.pdf

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What are the difference between the testes of a frog and a rat in terms of tunica albuginea, interstitial cells, seminiferous tubules, basement membrane, sertoli cells Solution STRUCTURES FROG RAT TUNICA ALBUGINEA Thick fibrous sheet of connective tissue. Composed of collagen fibers and occasional smooth muscle cells. INTERSTITIAL CELLS Endocrinal in nature and secrete mainly androsterone and testosterone Major source of testosterone and also participate in steroid biosynthesis. SEMINIFEROUS TUBULES Lined by single layer of cuboidal cells. These cuboidal cells form the germinal epithelium Germinal epithelial component is absent SERTOLI CELLS Involved In providing nutrients to mature spermatozoa Capable of steroid biosynthesis and secretion of inhibin and androgen binding protein STRUCTURES FROG RAT TUNICA ALBUGINEA Thick fibrous sheet of connective tissue. Composed of collagen fibers and occasional smooth muscle cells. INTERSTITIAL CELLS Endocrinal in nature and secrete mainly androsterone and testosterone Major source of testosterone and also participate in steroid biosynthesis. SEMINIFEROUS TUBULES Lined by single layer of cuboidal cells. These cuboidal cells form the germinal epithelium Germinal epithelial component is absent SERTOLI CELLS Involved In providing nutrients to mature spermatozoa Capable of steroid biosynthesis and secretion of inhibin and androgen binding protein.

What are the difference between the testes of a frog and a rat in te.pdf

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Which of the following occurs with disruptive selection a. Natural selection favors phenotypes of one extreme, resulting in a population mean shift over several generations b. Natural selection favors phenotypes of both extremes, resulting in an increase in intermediate phenotypes over time c. Natural selection favors phenotype values near the population mean, resulting in no shift in mean trait value over time and a more narrow bell curve d. Natural selection selects against trait values that are close to the population mean, resulting in a bimodal distribution of trait values Solution Disruptive selection happens when conditions favor individuals at both extremes of a phenotypic range over individuals with intermediate phenotypes. one example is a population of black- bellied seedcracker finches in Cameroon whose members show two distinctly different beak sizes. small-billed birds feed generally on soft seeds, whereas large-billed birds specialize in cracking hard seeds. it seems that birds with intermediate-sized bills are comparatively inefficient at cracking both types of seeds and thus have lower relative fitness..

Which of the following occurs with disruptive selectiona. Natural .pdf

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When comparing replicate populations of plants that are grown in the field or in controlled greenhouse conditions, the field population will likely have _ V_E, _ V_P, and heritability. [no written explanation necessary] higher/higher/higher higher/higher/lower higher/lower/higher higher/lower/lower lower/higher/higher lower/higher/lower lower/lower/higher lower/lower/lower Solution Correct answer: g Enivornmental variation (VE) is lower and Phenotypic Variation (VP)is lower, Heritability is higher.

When comparing replicate populations of plants that are grown in the .pdf

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What is wrong with the following definition of headInsert? struct Node { int item; Node* link; }; typedef Node* NodePtr; void headInsert(NodePtr& head, int data) { NodePtr tmp = new Node; tmp->item = data; head->next = tmp; tmp->next = head->next; } NodePtr head; headInsert(head, 4); A. tmp should be declared to be a Node not a NodePtr. B. head->next is pointing to NULL. C. Nothing is wrong. D. If there were any nodes following head they are now lost. Solution B. head->next is pointing to NULL. Because of which tmp is not getting stored in the linked list Correct Code: struct Node { int item; Node* link; }; typedef Node* NodePtr; void headInsert(NodePtr& head, int data) { NodePtr tmp = new Node; tmp->item = data; // next of tmp should be assigned head tmp->link = head; head = tmp; }.

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What is the BEST way to determine if there are any gaps in your thes.pdf

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What are two specialized epidermal cells and what do they do for the plant? Solution The two specialized epidermal cells in the plant are 1. Guard cells 2. Trichome. Guard cells are kidney shaped cells with rich green chloroplasts.Guard cells occur in form of pairs. A pair of guard cells and pore are referred as stomate. Guard cells perform two different functions. At daytime, guard cells open small pores in the plant epidermal surface allowing carbon dioxide to diffuse from atmosphere into the chlorenchyma tissue required for the photosynthesis process. It also regulates oxygen exchange. At night-time guard cells close the pores in the epidermal cells preventing water loss from leaves. Guard cells can alter the size of stomata by changing its structure in response to water exchange. Trichomes are branched hairy projections of modified epidermal cells. Trichomes are multi cellular with spine shape or hook shaped. Trichomes are branched or unbranched, unicellular or multicellular. Trichomes function to prevent water loss through evapouration and retains moisture. It secretes a sticky substance that repels insects..

What are two specialized epidermal cells and what do they do for the.pdf

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What are the basic differences and commonalties between Structured Analysis (process- oriented). Information Engineering (data-oriented), and Object-Oriented approaches for software development? Solution Structured Analysis is technique used in software development to analyse the designed system to its correctness. The analysis is made to see to what extent the designed system meets the requirement of the user. The structured analysis uses data flow diagrams to analyze the designed structure. It is believed that if the structure is good then the data will be good, but doesnot pay attention on data in detail. Data oriented analysis is used where data plays the utmost role. The diagrams that are used for analysing the data oriented design is ER diagrams. Data oriented design establishes the relations that are required for the development of the system on various data.The data analysis concentrates on incorporating additional features in process to ensure the data integrity. Object oriented analysis unlike of the above two concetrates on the objects, which are the instance of class that consists of both data and process or logic. The data oriented anaylsis cannot work on the structured process. similalry structures analysis cannot be performed on data oriented design.The process oriented and object oriented analysis are widely in use when compared to data oriented analysis..

What are the basic differences and commonalties between Structured An.pdf

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Use the following word bank to complete the number 51-100 This structure protects fish gills This Is the site In mammals where gases are exchanged with the environment This Is the site of gas exchange in insects This vein collects blood from the lower body and carries it to the heart The common opening of the digestive system, the excretory system and the reproductive system Is known as the Bile is produced in the Phloem, responsible for transporting nutrients in plants is composed of two types of cells and cells. These organs in spiders produce silk In this kind of circulatory system blood is carried entirely in vessels: This organ is both and endocrine and exocrine gland This group includes platypus and spiny anteaters. Solution 1. Operculum- Operculum of bony fish is a hard bony flap covering and protecting the gills. 2. Alveolus-Tiny sacs of the lungs which helps in gaseous exchange. 3. Tracheal Tube- In insects the gaseous exchange occurs through a network of tubes called the tracheal tube. 4. Inferior Vena Cava-It carries the blood from the lower body and carries it to the right atrium. 5. Cloaca-It is the common opening between the external environment and the digestive, reproductive and excretory system..

Use the following word bank to complete the number 51-100 This struc.pdf

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The nucleus is something we are debating on Celia\'s thread too. Although not called a nucleus, but rather a nucleoid in the prokaryotic cell, how does the organization of genetic material differ in the prokaryotic cell? Solution Ans. The genetic material of both eukaryotes and prokaryotes are chemically the same- being constituted of A, C, T G nucleotides forming double stranded DNA helix. However, their organization (from naked DNA to chromosome) differ in following aspects- I. Eukaryotic DNA is always linear, whereas that of prokaryotes is circular. II. Eukaryotic DNA is associated with basic histone proteins that facilitates its supercoiling to 30nm fiber. Prokaryotes lack histone proteins. Instead prokaryotic DNA is associated with various nucleoid- associated proteins, histone-like proteins and transcription factors that facilitate its condensation into loosely-packed loops. III. Eukaryotic genome is housed in a double membrane- bound organelle called the nucleus. Prokaryotes lack membrane bound organelles including nucleus. The regions of prokaryotic cytoplasm where genomic DNA is present in condensed form is called nucleoid..

The nucleus is something we are debating on Celias thread too. Alt.pdf

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19. Photosynthetic organisms are found in the Domain Eukarya as well as Bacteria. Which ofthe following is NOT a photosynthetic feature common to BOTH? A. Chloroplasts are the site of photosynthesis B. Photosynthetic ments are embedded in membranes C. Pigment molecules transfer energy to a reaction center molecule D. ns are transferred to produce ATP E. LJ Conversion of light energy into Chemical energy occurs. Solution B. Photosynthetic pigments are embedded in membranes. Rest all are correct..

19. Photosynthetic organisms are found in the Domain Eukarya as well .pdf

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Qualitative methods tend to have those characteristics. Established methods, open-ended questions, interview data/observation data, themes & patterns Emerging methods, open-ended questions, observations data/observation data, statistical patterns Emerge methods, open-ended questions, observes data/observation data, themes & patterns Emerging methods, open-ended quests, survey data/observation data, themes & patterns. Solution Answers: 4. (d) Emerging methods, open-ended questions, survey data/observational data, themes and patterns. Qualitative research involves emerging trends/methods containing open ended questions. The data is collected through surveys and observations, followed by thematic analysis. It emphasizes examining and recording patterns. 5. False Faculty are comfortable with systematic procedures and transformative approaches to inquiry. The principal way of new knowledge development is what is often called \"systematic inquiry\", i.e., a careful, deliberate effort to deal with a problem, to investigate something inadequately known or understood. The term Transformative Inquiry (TI) refers to a multi-faceted process that includes particular concepts, actions and philosophical approaches..

Qualitative methods tend to have those characteristics. Establis.pdf

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onchange, onclick and onkeyup, are some examples of ...A. Tags B. .pdf

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1. My anatomy: I am quite small as compared to many microorganisms. Members within my family are typically too small to be observed with an optical microscope. My chemical composition includes relatively few proteins and a lipid bilayer. 2. My growth and metabolism: I am an obligate parasite relying on a cellular host to reproduce. Independent of my host, I have no metabolism. 3. My genetic makeup: whether it is in the absence or in the presence of my host, the genetic information which I use to replicate myself is never in the form of DNA. As a free entity, my genetic information is encoded by a coding RNA (positive). However, I can also generate non-coding RNA during my life cycle within the cell. 4. I am fairly restricted in the hosts in which I am pathogenic, being limited to primates such as humans. 5. The symptoms associated with the disease that I cause include feeling tired, fever, weakness, headache, and a sore throat. In more severe cases, hosts will bleed from their mouths and their eyes as well as vomit blood. 6. My origins are thought to be in Africa from where I spread to the Americas through the slave trade. 7. I am not directly transmitted from primate to primate. I require a vector, which like a good acquaintance of mine the malaria causing pathogen, is the mosquito. 8. The first report of an epidemic in the Americas for which I am held responsible occurred in 1648 where I am held accountable for destroying the Yucatan Mayans. Solution The disease described here is yellow fever, which is caused by the yellow fever virus, arbovirus of genus flavivirus. The vector is mosquito Aedes aegypti..

1. My anatomy I am quite small as compared to many microorganisms. M.pdf

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Normalization work as through a series of stages called normal forms Normalization is a process that is used for changing attributes to entitles. In order to meet performance requirements you may have to denormalize portions design. Denormalization produces a lower normal form. Normalization is a very important database design ingredient and the highest level is always the desirable. A table is in BCNF if every determinant in the table is a candidate key. A table is in fourth normal form if it is in third normal form and has no independent multivalued dependencies. Repeating groups must be eliminated by making sure that each row defines a single entity. It is possible for a table in 2NF to exhibit transitive dependency, where one or more nonprime attributes functionally determine other nonprime attributes. A determinant is any attribute whose value determines other values within a column. A dependency of one nonprime attribute on another nonprime attribute is a partial dependency. Attributes should clearly define participation, connectivity, and document cardinality. An attribute that is part of a key is known as a(n) attribute. important nonprime prime entity Data redundancy produces. slower lookups robust design efficient storage use data integrity problems Solution 1) Normalization works through a series of stages called normal forms. Answer :- True 2) Normalization is a process that is used for changing attributes to entities. Answer :- False 3) In order to meet performance requirements, you may have to denormalize portions of the database design. Answer :- True 4) Denormalization produces a lower normal form. Answer :- True 5) Normalization is a very important database design ingredient and the highest level is always the most desirable. Answer :- False 6) A table is in BCNF if every determinant in the table is a candidate key. Answer :- True 7) A table is in fourth normal form if it is in third normal form and has no independent multivalued dependencies. Answer :- True 8) Repeating groups must be eliminated by making sure that each row defines a single entity. Answer :- True 9) It is possible for a table in 2NF to exhibit transitive dependency, where the primary key may rely on one or more nonprime attributes to functionally determine other nonprime attributes. Answer :- True 10) A determinant is any attribute whose value determines other values within a column. Answer :- False 11) A dependency of one nonprime attribute on another nonprime attribute is a partial dependency. Answer :- False 12) Attributes should clearly define participation, connectivity, and document cardinality. Answer :- False.

Normalization work as through a series of stages called normal forms .pdf

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