What are the benefits of endocrine signaling in comparison to neuron.pdf

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What are the benefits of endocrine signaling in comparison to neuronal signaling? Solution The endocrine signalling is long lasting,broad impact and is difficult to reverse.The endocrine system relies on hormones to elicit responses from target cells. The nervous system conveys high-speed electrical signals along specialized cells called neurons,these signals regulate other cells.The response to stimuli by the nervous system is near instantaneous,although the effects are often short lived..

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Discuss the benefits of event-driven programming. Contrast event-driven programming with procedural programming. Discuss how messages from other programs can be viewed as events. Give examples. Give examples of tasks that require the program to guide the user. Should programs be purely event driven, procedural, or a mixture of both? Why? Solution The benefit of event-driven programming is that it\'s very intuitive and naturally well-suited to applications whose control flow are based, not on its structure, but rather on internal or external events. This is why event-driven programming is so popular with GUI programming. Likweise, SOA and client-server architectures are naturally event-driven as well. We if we find contrast between Procedural and event-based then applications written in procedural languages execute by proceeding logically through the program code, one line at a time. Logic flow can be temporarily transferred to other parts of the program through the GoTo, GoSub, and Call statements, directing the program from beginning to end. In contrast, program statements in an event-driven application execute only when a specific event calls a section of code assigned to that event. Events can be triggered by keyboard input, mouse actions, the operating system, or code in the application. For example, consider what happens when the user clicks a command button named Command1 on a form. The mouse click is an event. When the Click event occurs, Visual Basic executes the code in the Sub procedure named Command1_Click. When the code has finished running, Visual Basic waits for the next event. Messages from other programs/threads can be seen as events, as they trigger some instance of code to execute..

Discuss the benefits of event-driven programming.Contrast event-dr.pdf

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Before 1900, despite its weaknesses in effective management of workers, manufacturing leadership was well provided by top management. They were technological entrepreneurs, archictects of productive systems, veritable lions of industry. But when they delegated their production responsibilities to a second-level department, the factory institution never recovered its vitality. The lion was tamed. It\'s management systems became protective and generally were neither enterpreneual nor strategic. Production managers since then have typically had little to do with initiating substantially new process technology-in contrast to their predecessors before 1900 (skinner 1985). D) how is Japan (or Germany) different from (or the same as) America with regards to this trend in manufacturing leadership? E) taking the structural charestaristics of manufacturing enterprises (e.g., scale, complexity, pace of technological change) as given, what can be done to revitalize manufacturing leadership? Solution Strategic Windows: their nature. The nature and purpose of strategy and how it is formulated. The nature of marketing strategy and how this should take account of the interests of various stakeholders when involving such things as, product/service development and delivery, promotional mix, support services, manufacturing and production processes, R&D, and material purchasing affect the stakeholders. Other factors in the business environment that influence marketing strategy: political, economic, socio-cultural and technological (PEST). Marketing and competitors: how a firm must be able to position itself competitively in the minds of its customers so that its products and services stand out very favourably in important respects in relationship to competitors. Matching the firm’s products / services with opportunities and threats in the market place. The limited periods during which the fit between the key requirements of a market and the particular competencies of a firm competing in that market are at an optimum. Investment in a product line or a market area should be timed to coincide with periods during which a strategic window is open. Correspondingly, withdrawal should be considered where something which was a good fit, is no longer a good fit. Ways in which a market can evolve and how firms might develop a competitive strategy to take advantage of Strategic Windows. Portfolio Analysis How organisations create their own environments rather than simply adapt to existing ones. How they select the strategic windows of opportunities and threats through which they want to look out into the world and develop and market product and services to meet the needs of what they observe to be required in the face of environmental turbulence. How well the fit between an organization’s products/services meet the needs presented by the windows of opportunities and threats is a fitting start for exploring the subject of strategic marketing. It introduces the many factors t.

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Alpha particle radiation sensor. A silicon diode radiation sensor is used to check the radioactive source in a smoke detector. According to the label, the source is given as 1 music. The source is Ain-241 with an energy of 486 times 10^6 e V and an activity of 195 times 10^10 B q. The energy needed to generate an electron-hole pair is 61 e V. The diode is connected to an inverse potential of 12 V, has a thickness of 0.5 mm, and the mobility\'s are 450 cm^2/V s for holes and 1350 cm^2/V s for electrons. The radiation from Ain-241 is mostly a particles that penetrate very little into the silicon and hence may be assumed to be absorbed at the surface of the diode For this reason the junction is exposed directly so radiation does not pass through the electrodes. What is the current through the diode if all particles emitted are trapped by the diode? Solution Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their production in standard alpharadioactive decay, alpha particles generally have a kinetic energy of about 5 MeV, and a velocity in the vicinity of 5% the speed of light. (See discussion below for the limits of these figures in alpha decay.) They are a highly ionizing form of particle radiation, and (when resulting from radioactive alpha decay) have low penetration depth. They are able to be stopped by a few centimeters of air, or by the skin. However, so-called long range alpha particles from ternary fission are three times as energetic, and penetrate three times as far. As noted, the helium nuclei that form 10–12% of cosmic rays are also usually of much higher energy than those produced by nuclear decay processes, and are thus capable of being highly penetrating and able to traverse the human body and also many meters of dense solid shielding, depending on their energy. To a lesser extent, this is also true of very high-energy helium nuclei produced by particle accelerators..

Alpha particle radiation sensor. A silicon diode radiation sensor is .pdf

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A ship was caught in a storm and driven to a small island in the middle of the South Pacific. For generations, no one encountered this small colony, who despite the odds made a thriving settlement on their island. After many years, what characteristics would be unique about this population? A: The people on the island would no longer be considered humans after evolving independently for many generations. B: The people on the island would all look very different from each other, and they may exhibit abnormally high occurrences of traits that are rare in the general human population. C: The people on the island would all look very different from each other, and they may have evolved some odd characteristics that aren\'t seen in other humans. D: The people on the island would all look very similar to each other, and they may exhibit abnormally high occurrences of traits that are rare in the general human population. Solution There is a small group of population who lived seperately from the human for many many years . In this case, the small population would have bred amongst themselves are continued their colony. This small group say after a thousand years look more like each other than any other human on the earth as they had been isolated from the rest of the world. So the answer should be D)The people on the island would all look very similar to each other, and they may exhibit abnormally high occurrences of traits that are rare in the general human population.

A ship was caught in a storm and driven to a small island in the mid.pdf

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8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its reactions? 1. 2. 3. 4. 10. What is the BASIC difference between substrate-level phosphorylation and oxidative phosphorylation? (do not tell me which pathways are involved). 11. Exactly….and I mean exactly…..which chemical reaction in the aerobic metabolic pathways, actually uses oxygen, and what does it do to the oxygen that it uses. Solution 8). Enzymes are the proteins, which decrease the activation energy of reaction and catalyse the reaction process. All enzymes are catalysts because, Cells decrease the rate of reaction by 1). Altering pH 2). Accumulation of products 3). Altered temperature 4). Inactivation of catalyst.

8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its rea.pdf

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As leaf temperature increases from 20°C to 35°C, the quantum yield of C3 photosynthesis is directly affected by The decreasing ratio CO2 to O2 in the gas phase The increased reaction rates of photorespiration The decreasing ratio CO2 to O2 in the aqueous phase Degradation of the thylakoid membrane and proteins RubisCO’s improved affinity for O2. Solution Answer is : The decreasing ratio CO2 to O2 in the aqueous phase. Reason: When the leaves are hot, CO2 becomes less soluble in the aqueous phase faster than O2, and the CO2/O2 ratio drops.

As leaf temperature increases from 20°C to 35°C, the quantum yield o.pdf

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Among 10 people traveling in a group, 2 have outdated passports. It is known that inspectors will check the passports of 20% of the people in any group passing their desks. The group can go as a whole (all 10) to one desk or can split into two groups of 5 and use two different desks. How should members of the group arrange themselves to maximize the probability of getting by the inspectors without having the outdated passports detected? I do not know how to figure out each probability... The section is on counting rules - i am guessing it is without replacement and order does not matter. Thank you for your help Solution Inspectors would check 20% of people in the group Hence if 10 people, 2 would be checked or if 5 one person would be checked If 10 people go to one desk, p for having outdated passport = 0.2 X - no of people having correct passports in 10 is binomial with p = 0.8 They would be caught only if Inspector selects any one person from the two. Y - no of people having outdated passports and p for Y =0.2 P(Y>=1) = 1-P(Y=0) = 1-(0.8)10 = 0.8926 --------------------------------------------- If goes into two groups then either each group can have one outdated or one group no outdated and other groups 2 outdated. For each group one outdated prob of being caught = P(y=1) in I group + P(y=1 in second group)-P( P(y=1) in I group *P(y=1 in second group) as these two events are independent = {1-(0.8)5}+{1-(0.8)5}-{1-(0.8)5}{1-(0.8)5} =2(0,67232)-0.672322 = 0.8926 (same as the first case) ---------------------------------------------- If one group has both outdated and other has none then prob of being not caught = 1(Prob of atleast one in the group having outdated 2) Here p for good passport =0.6 = P(Y>=1) = 1-P(y=0) = 1-(0.6)5 =0.92224 Hence this is the best method to go. Divide into two groups with 2 outdated in one group.

Among 10 people traveling in a group, 2 have outdated passports. It .pdf

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1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are available. Similarly, wings are normally long, but there are strains with short wings. A female from a pure line with brown eyes and short wings is crossed with a male from a normal pure line. The F1 consists of normal females and short-winged males. An F2 is then produced by intercrossing the F1. Both sexes of F2 flies show phenotypes as follows: 3/8 red eyes, long wings 3/8 red eyes, short wings 1/8 brown eyes, long wings 1/8 brown eyes, short wings Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own invention. State the genotypes of all three generations and the genotypic proportions of the F1 and F2. 2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs divides mitotically, what are the genotypes of the daughter cells? 3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ; b = brown body; F = forked bristles; f = unforked bristles.) The results are black, forked 230 brown, forked 240 black, unforked 210 brown, unforked 250 Use the 2 test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid. 4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in each sex? (Invent and define your own gene symbols.) Solution Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will give rise to daughter cells with the genotype A/a; B/b; C/c. However, if the cell was to divide meiotically (reductional division), the genetic constitution would be different as only one half of the parental chromosomes goes to the daughter cell. So the genotypes would be as follows: ABC or AbC or ABc or Abc or aBC or abC or aBc or abc Answer 3. For the given cross BbFf X bbff For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf). For dihybrid cross, the degree of freedom will be 4 – 1 = 3 Chi square test is as follows BF Bf bF bf Totals Observed phenotype 230 210 240 250 930 Expected phenotype 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930x ¼= 232.5 930 Observed – expected -2.5 -22.5 7.5 17.5 (Obs – Exp)2 Expected 6.25/232.5= 0.0268 506.25/232.5= 2.1774 56.25/232.5= 0.2419 306.25/232.5= 1.3172 3.7633 Looking at the probability value table for degree of freedom = 3, df .90 .70 .50 .30 .20 .10 .05 .01 3 0.58 1.42 2.37 3.67 4.64 6.25 7.82 11.35 Hence we can deduce that there is more than 20% probability for the occurrence of this event. Answer 4. Let XD = Normal phenotype (tall) Xd = Dwarf phenotype P = Pink coat p = Br.

1.) The normal eye color of Drosophila is red, but strains in which .pdf

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29. Code an application program that keeps track of student information at your college. Include their names, identification numbers, and grade point averages in a fully encapsulated, homogeneous sorted singly linked list structure. When launched, the user will be asked to input the initial number of students and the initial data set. Once this is complete, the user will be presented with the following menu: Enter: 1 to insert a new student\'s information, 2 to fetch and output a student\'s information, 3 to delete a student\'s information, 4 to update a student\'s information, 5 to output all the student information in sorted order, and 6 to exit the program. The program should perform an unlimited number of operations until the user enters a 6 to exit the program. If the user requests an operation on a node not in the structure, the program output should be “node not in structure.” Otherwise, the message “operation complete” should be output. in java Solution Dear Asker, Following 3 classes consist the colution to this problem. This will give you an idea to approach this problem: //Student Class package studentinfo; public class Student implements Comparable { private String name; private String id; private double gpa; public Student(String id, String name, double gpa) { this.name = name; this.id = id; this.gpa = gpa; } /** * @return the name */ public String getName() { return name; } /** * @param name the name to set */ public void setName(String name) { this.name = name; } /** * @return the id */ public String getId() { return id; } /** * @param id the id to set */ public void setId(String id) { this.id = id; } /** * @return the gpa */ public double getGpa() { return gpa; } /** * @param gpa the gpa to set */ public void setGpa(double gpa) { this.gpa = gpa; } @Override public int compareTo(Student o) { if (this.id.compareTo(o.id) < 0) { return -1; } else if (this.id.compareTo(o.id) > 0) { return 1; } else if (this.gpa < o.gpa) { return -1; } else if (this.gpa > o.gpa) { return 1; } return 0; } public boolean equals(Student o) { if (this.name == o.name && this.id == o.id && this.gpa == o.gpa) { return true; } return false; } } //Student container class package studentinfo; import java.util.Collections; import java.util.LinkedList; public class StudentContainer { LinkedList studentList; public StudentContainer() { this.studentList = new LinkedList(); } public boolean hasStudent(String id, String name, double gpa) { return this.studentList.contains(new Student(id, name, gpa)); } public boolean insertStudent(String id, String name, double gpa) { if (!this.isValidString(id) || !this.isValidString(name) || !this.isValidGpa(gpa)) { return false; } this.studentList.add(new Student(id, name, gpa)); return true; } public boolean deleteStudentById(String id) { Student toRemove = null; for (Student e : studentList) { if (e.getId() == id) { toRemove = e; break; } } if (toRemove != null) { this.studentList.remove(toRemove); return true; } return false;.

29. Code an application program that keeps track of student informat.pdf

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Write the code for a small function called myStack, which creates a Stack object of String elements, named cities. Add \"Sydney\", then \"London\", then \"New York\" and then \"Beijing\" to cities. Then print each element on the screen, one per line. Print each element in a way that causes cities to be empty (i.e. have 0 elements) afterwards. (6 mks) Solution PROGRAM CODE: package stack; import java.util.Stack; public class MyStackClass { //stack function public static void myStack() { //creating a stack object called cities Stack cities = new Stack(); //Adding cities into the stack cities.push(\"Sydney\"); cities.push(\"London\"); cities.push(\"New York\"); cities.push(\"Beijing\"); //using while loop to loop through the stack till its empty while(!cities.isEmpty()) { // pop function is used to remove elements from the stack // pop() returns the elemets which is printed on the csreen System.out.println(cities.pop()); } } // Main function for the class public static void main(String args[]) { myStack(); // function call to the stack method } } OUTPUT: Beijing New York London Sydney.

Write the code for a small function called myStack, which creates a .pdf

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Why does the neutralization of an acid by a base often produce water? Solution Neutralisation refers to a reaction between an acid and a Base. This reaction often results in the production of water. It is due to a fact that an acid liberates H+ ions while as a Base releases OH- ions. These ions combine to form water. For example: HCl(aq) H+(aq) + Cl(aq) NaOH(aq) Na+(aq) + OH(aq) HCl + NaOH NaCl + H2O..

Why does the neutralization of an acid by a base often produce water.pdf

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Which statement about phloem transport is false? It takes place in sieve tubes It requires metabolic energy A high pressure potential is maintained in the sieve tubes At sinks, solutes are actively transported into sieve tube elements The mycorrhizal relationship allows a plant’s roots to ?? increase the amount of soil than can be accessed for nutrients. grow in nutrient-poor soils. avoid direct interaction with the soil. None of the above Solution 1. At sinks, solutes are actively transported into sieve tube elements. (False) Phloem is a vascular tissue in plants that actively transports synthesized food from the source tissue ( photosynthetic) to sink tissue (non-photosynthetic). The sieve tube elements (STEs) are the phloem tissue compartments which help in the transportation of food from source to sink. Phloem transport requires metabolic energy and a high-pressure potential is maintained at the sieve tubes. 2. Increase the amount of soil that can be accessed for nutrients Mycorrhiza is the symbiotic relationship between fungi and the plant roots on which they extensively grow. The network of fungal growth or mycorrhiza helps in increasing the surface area of the roots so that they can access more amount of soil and take up nutrients essential for plant growth..

Which statement about phloem transport is falseIt takes place in .pdf

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Which of the following is not a primitive data type Which of the following is not a primitive data type Solution Hi, PLease find primitive data type list in C: The primitive data types in c language are the inbuilt data types provided by the c language itself. Thus, all c compilers provide support for these data types. The following primitive data types in c are available: Integer Data Type int , short, long Float data Type float Double Data Type double Character Data Type char Void Data Type void.

Which of the following is not a primitive data type Which of th.pdf

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What are the three types of column chromatography and the principles behind how they work? For the most specificity/highest purification which type would you expect to achieve this. Solution Three type of column chromatography are as follows: 1.Gel permeation chromatograhy 2.Ion exchange chromatography 3.Adsorption Chromatography GEL PERMEATION CHROMATOGRAPHY his chromatographic technique for the separation of molecules on the basis of their molecular size and shape exploits the molecular sieve properties of a variety of porous materials. The terms exclusion or permeation chromatography or gel filtration describe all molecular separation processes using molecular sieves. The general principle of exclusion chromatography is quite simple. A column of microparticulate cross-linked copolymersgenerally of either styreneor divinylbenzene andwith a narrow range of pore sizes is in equilibrium with a suitable mobile phase for the analytes to be separated. Large analytes that are completely excluded from the pores will pass through the interstitial spaces between the particles and will appear first in the eluate. Smaller analytes will be distributed between the mobile phase inside and outside the particles and will therefore pass through the column at a slower rate, hence appearing last in the eluate ION EXCHANGE CHROMATOGRAPHY Separation of molecules that contain electrically charged groups is often accomplished best by ion exchange chromatography.The technique depends upon interactions between the charged groups of the molecules being separated and fixed ionic groups on an immobile matrix. Separation depends upon small differences in pK a values and net charges and upon varying interactions of nonpolar parts of the molecules being separated with the matrix. Since changes in pH can affect both the charges on the molecules being separated and those of the ion exchange material, the affinities of the molecules being separated are strongly dependent on pH. For example, proteins and most amino acids are held tightly by cation exchangers at low pH but not at all at high pH.Aqueous solutions are usually employed and the columns are packed with beads of ion exchange resins, porous materials containing bound ionic groups such as – SO 3 – , – COO – , – NH 3+ , or quaternary nitrogen atoms. Synthetic resins based on a crosslinked polystyrene are usually employed for separation of small molecules. For larger molecules chemical derivatives of cellulose or of crosslinked dextrans (Sephadex), agarose, or polyacrylamide are more appropriate. Positively charged ions, such as amino acids in a low pH solution, are placed on a cation exchange resin such as Dowex 50, which contains dissociated sulfonic acid groups as well as counter ions such as Na + , K + , or H + . The adsorbed amino acids are usually eluted with buffers of increasing pH containing sodium or lithium ions.Ion exchange chromatography of proteins and peptides is often done with such ion exchange materi.

What are the three types of column chromatography and the principles.pdf

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What does this statement mean...genes code for gene products? What does this statement mean...genes code for gene products? Solution Gene codes for gene product.The statement means gene expression which is a process by which a gene\'s information is used for the synthesis of a gene product that is functional.The gene products are proteins in case of protein coding genes while in case of non protein coding genes i,e t-RNA or snRNA , the gene product is RNA that is functional..

What does this statement mean...genes code for gene products W.pdf

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What a program to toggle PD3, PH7, and PC5 continuously without, disturbing the rest of the bits. Write a program to monitor bit PC3. When it is HIGH, send 0 times 55 to PORTD. Solution //Accessing Ports as SFRs using sfr data type sfr PD3=0x80; sfr PD7=0x90; sfr PC5=0xA0; void MSDelay(unsigned int); void main(void) { while (1) { PD3=0x55; PD7=0x55; PC5=0x55; MSDelay(0); PD3=0xAA; PD7=0xAA; PC5=0xAA; MSDelay(0); } }.

What a program to toggle PD3, PH7, and PC5 continuously without, dist.pdf

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True of False results that are practically significant will always .pdf

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The waiting time T between successive occurrences of an event E in a discrete-time renewal process has the probability distribution P(T = 1) = 0.7, P(T = 2)- 0.3. The waiting time to the sixth occurrence of E is denoted by W6. Find the probability P(W6 = 10). 0.0595 0.0494 0.0393 Solution Let the waiting time to 1st occurence be T1 , between occurence 1 and 2 be T2 , and.... between occurence 5 and 6 be T6. P(T1 + T2 + T3 + T4 + T5 + T6 = 10) : For the 6 time intervals to add up to 10, as the only possible interval values are 1 and 2, 4 of the intervals have to have the value 2 and remaining 2 have to have the value 1 2+2+2+2+1+1 = 10 These values can be assigned to 6 intervals in : 6C4 * 2C2 ways = 15 ways P(T1 + T2 + T3 + T4 + T5 + T6 = 10) = 15 * [P(T=1)]2 * [P(T=2)]4 = 15*0.72 *0.34 = 0.0595 Answer: 0.0595.

The waiting time T between successive occurrences of an event E in a .pdf

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The half-life of Carbon-14 is about 5700 years. You have found a fossil in sedimentary rocks that have 1/128 the normal (modern) expected ratio of Carbon-14C. Based on this information, you estimate that it is about _______ years old. 44 5700 39,900 729,600 Solution The fossil in 39,900 years old. At 1/128 the carbon-14 of living organisms, it has gone through seven half-lives of carbon-14 (5700 years x 7 = 39,900 years).

The half-life of Carbon-14 is about 5700 years. You have found a fos.pdf

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The (circle one meanmedian) would be a better measure of central te.pdf

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