Answer As long as the two substituents on any one of the two carbon.pdf

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Answer: As long as the two substituents on any one of the two carbons are not the same, it would be capable of cis/trans isomerisim. ===== Explanation: In your example of A--C=C--A, since each of the carbons have an A group and a hydrogen (not shown), it is possible for isomerism (A group is different from H). In your example of B--C=C--A, it is also capable of isomerism because the two groups on each carbon are different (B group is different from H on the first carbon, A group is also different from H on the second carbon). However, if for example, you have a two B groups both attached to the left carbon (without any hydrogens), it cannot be either cis/trans. ***Do note however, that if you have more than two different groups across the double bond overall (ex. Group A, B, and C), you should use the E/Z naming system (organic chemistry) instead of cis/trans. Solution Answer: As long as the two substituents on any one of the two carbons are not the same, it would be capable of cis/trans isomerisim. ===== Explanation: In your example of A--C=C--A, since each of the carbons have an A group and a hydrogen (not shown), it is possible for isomerism (A group is different from H). In your example of B--C=C--A, it is also capable of isomerism because the two groups on each carbon are different (B group is different from H on the first carbon, A group is also different from H on the second carbon). However, if for example, you have a two B groups both attached to the left carbon (without any hydrogens), it cannot be either cis/trans. ***Do note however, that if you have more than two different groups across the double bond overall (ex. Group A, B, and C), you should use the E/Z naming system (organic chemistry) instead of cis/trans..

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hydrogens), it cannot be either cis/trans.
***Do note however, that if you have more than two different groups across the double bond
overall (ex. Group A, B, and C), you should use the E/Z naming system (organic chemistry)
instead of cis/trans.

Light is a form of energy that can be released by an atom. It is made up of many small particle-like packets that have energy and momentum but no mass. These particles, called light photons, are the most basic units of light. Atoms release light photons when their electrons become excited. When an atom gains or loses energy, the change is expressed by the movement of electrons. When something passes energy on to an atom, an electron may be temporarily boosted to a higher orbital (farther away from the nucleus). The electron only holds this position for a tiny fraction of a second; almost immediately, it is drawn back toward the nucleus, to its original orbital. As it returns to its original orbital, the electron releases the extra energy in the form of a photon, in some cases a light photon. Solution Light is a form of energy that can be released by an atom. It is made up of many small particle-like packets that have energy and momentum but no mass. These particles, called light photons, are the most basic units of light. Atoms release light photons when their electrons become excited. When an atom gains or loses energy, the change is expressed by the movement of electrons. When something passes energy on to an atom, an electron may be temporarily boosted to a higher orbital (farther away from the nucleus). The electron only holds this position for a tiny fraction of a second; almost immediately, it is drawn back toward the nucleus, to its original orbital. As it returns to its original orbital, the electron releases the extra energy in the form of a photon, in some cases a light photon..

Light is a form of energy that can be released by.pdf

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1) Presence of jaws with paired fins in fish helps to feed them. Fins helps them with increased stability and maneuverability and jaws helps to capture the prey and seize it. 2) Fish with jaws need to regulate their body temperature so they need to feed themselves often. It capture the prey and send it to stomach for digestion and for temporary storage. Jawless fish donot regulate their own body temperature so they don\'t need temporary storage like stomach and also they have sucking mouth and sharp teeth which helps in digestion without the need for pre digestion with gastric juice. So the fish which needs more food has jaw (Capture large prey) with stomach (Temporary storage and pre digestion). The fish which needs less food has no jaw and no stomach but with sucking mouth and also they have no fins. Jawless fish examples : Lampreys and hagfish. Majority of the fishes are with jaws. 3) Two chambered heart have one atrium and one ventricle and three chambered heart have two atrium and one ventricle. There are two types of circulatory system that is single circulation system (fish) and double circulation system (reptiles, mammals) In single circulation system, blood from the parts of the body enters heart through atrium and then ventricle pumps the blood to gills where re-oxygenation takes place. From the gills, blood is sent back to body parts. Blood enters the heart only once that is only deoxygenated blood enters. In double circulation system, blood enters the heart twice that is from parts of the body to heart then from heart to pulmonary circulation (Lungs) and from lungs to heart then heart to parts of the body. Both deoxygenated and oxygenated blood enters the heart. Solution 1) Presence of jaws with paired fins in fish helps to feed them. Fins helps them with increased stability and maneuverability and jaws helps to capture the prey and seize it. 2) Fish with jaws need to regulate their body temperature so they need to feed themselves often. It capture the prey and send it to stomach for digestion and for temporary storage. Jawless fish donot regulate their own body temperature so they don\'t need temporary storage like stomach and also they have sucking mouth and sharp teeth which helps in digestion without the need for pre digestion with gastric juice. So the fish which needs more food has jaw (Capture large prey) with stomach (Temporary storage and pre digestion). The fish which needs less food has no jaw and no stomach but with sucking mouth and also they have no fins. Jawless fish examples : Lampreys and hagfish. Majority of the fishes are with jaws. 3) Two chambered heart have one atrium and one ventricle and three chambered heart have two atrium and one ventricle. There are two types of circulatory system that is single circulation system (fish) and double circulation system (reptiles, mammals) In single circulation system, blood from the parts of the body enters heart through atrium and then ventricle pumps the blood to gills .

1) Presence of jaws with paired fins in fish helps to feed them. Fin.pdf

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You\'re looking for Ka, the dissociation constant, so you measure this by adding the substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is 4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0, there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of products/concentration of reactants. always look at the coefficients infront of the molecules in the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be 1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final concentration is 0.063-concentration of hydronium...but since the concentration of the hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E- 5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please rate. Solution You\'re looking for Ka, the dissociation constant, so you measure this by adding the substance to water. 1. Write the balanced equation HBrO + H2O = BrO- + H30+ 2. the pH is 4.95, so you know its acidic and there has to have hydronium as a product. if the pH was >7.0, there would be OH- instead. the pH is a measure of H ions, and to get the concentration, you simply do 10^(-pH). 10^(-4.95) = 1.12E-5 3. The equation for Ka is concentration of products/concentration of reactants. always look at the coefficients infront of the molecules in the reaction. in this case, it is [BrO-)][H3O+]/[HBrO] *liquids and solids are never included in this, only aqueous solutions and gases are* 4. Your concentration of H3O was found to be 1.12E-5, and because the H30 and BrO are in a 1:1 ration in the equation, the concentration for BrO is also 1.12E-5. The concentration for HBrO as a reactant is given, but the final concentration is 0.063-concentration of hydronium...but since the concentration of the hydronium is so small, it doesnt at all affect the final answer. 5. your equation for Ka=((1.12E- 5)(1.12E-5))/(0.063-1.12E-5). The final answer for your Ka is 2.0E-9 Hope this helps, please rate..

Youre looking for Ka, the dissociation constant.pdf

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Okay so dipole-dipole interactions would occur between polar molecules such as water. The H has a + charge while O has a - charge. Dipole- induced dipole(or dispersion forces) would be like a molecule that has a polar part to it such as a -COOH group and then a nonpolar part such as a long hydrocarbon chain. The nonpolar part has no charge except the transient charge among the different atoms. I hope that helps Solution Okay so dipole-dipole interactions would occur between polar molecules such as water. The H has a + charge while O has a - charge. Dipole- induced dipole(or dispersion forces) would be like a molecule that has a polar part to it such as a -COOH group and then a nonpolar part such as a long hydrocarbon chain. The nonpolar part has no charge except the transient charge among the different atoms. I hope that helps.

Okay so dipole-dipole interactions would occur be.pdf

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remote operating system installation Remote Installation Services (RIS) ships as a part of the Windows® 2000 Server software system. This document outlines the steps necessary to put in, configure, and use RIS. RIS was designed to cut back the prices incurred by pre-installing or physically visiting every consumer pc to put in the software system (OS). By combining RIS with alternative Windows 2000 IntelliMirror™ management technologies features—User knowledge Management, software package Installation and Maintenance, and User Settings Management—companies get pleasure from higher disaster recovery with easier OS and application management. Prerequisites See Appendix A below to confirm that each your server and consumer hardware meet the remote installation hardware necessities. The drive on the server wherever you decide on to put in RIS should be formatted with the NTFS filing system. RIS needs a big quantity of space and can\'t be put in on constant drive or partition on that Windows 2000 Server is put in. make sure that the chosen drive contains enough free space for a minimum of one full Windows 2000 skilled compact disc—a minimum of roughly 800 megabytes (MB)–1 GB (GB). RIS needs many alternative services that conjointly ship as a part of Windows 2000 Server. These services will be put in on individual servers, or all on one server, counting on your network design: Domain Name Service (DNS) Server. Remote installation depends on DNS for locating the directory service and consumer machine accounts. you\'ll use any Windows 2000 Active DirectoryTM service-compliant DNS server, otherwise you will use the DNS server given Windows 2000 Server. Dynamic Host Configuration Protocol (DHCP) Server. RIS needs a DHCP server to be gift and active on the network. The remote boot-enabled consumer computers receive AN information processing address from the DHCP server before contacting RIS. Active Directory. RIS depends on Windows 2000 Active Directory for locating existing consumer machines further as existing RIS servers. RIS should be put in on a Windows 2000- based server that has access to Active Directory. this will be a site controller or a server that\'s a member of a site with access to the Active Directory. To ensure a in installation, you want to install and tack together the extra services delineate higher than so as for RIS to perform. additionally, certify that you just have each the Windows 2000 Server and skilled CDs on the market. Before starting this gradual guide, you wish to create the common infrastructure, that specifies a specific hardware and software package configuration. The common infrastructure is roofed within the Common Infrastructure gradual guide, \"Part 1: putting in a Windows 2000 Server as a site Controller\" http://www.microsoft.com/windows2000/techinfo/planning/server/serversteps.asp. If you\'re not victimization the common infrastructure, you wish to create the suitable changes to the current instruction set. Alth.

remote operating system installationRemote Installation Services (.pdf

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Title of this process is \"The Project Life Cycle (Phases)\" The project manager and project team have one shared goal: to carry out the work of the project for the purpose of meeting the project’s objectives. Every project has a beginning, a middle period during which activities move the project toward completion, and an ending. A standard project typically has the following four major phases: initiation, planning, implementation, and closure. Taken together, these phases represent the path a project takes from the beginning to its end and are generally referred to as the project “life cycle.” Initiation Phase During the first phases, the initiation phase, the project objective is identified; this can be a business problem or opportunity. An appropriate response to the objective of the project is documented with recommended solution options. A feasibility study is conducted to investigate whether each option addresses the project objective and a final recommended solution is determined. Issues of feasibility (“can we do the project?”) and justification (“should we do the project?”) are addressed. Once the recommended solution is approved, a project is initiated to deliver the approved solution and a project manager is appointed. Approval is then sought by the project manager to move onto the detailed planning phase. Planning Phase The planning phase, is where the project solution is further developed in as much detail as possible and the steps necessary to meet the project’s objective are planned. In this step, the team identifies all of the work to be done. The project’s tasks and resource requirements are identified, along with the strategy for producing them. A project plan is created outlining the activities, tasks, dependencies, and timeframes. The project manager coordinates the preparation of a project budget by providing cost estimates for the labor, equipment, and materials costs. The budget is used to monitor and control cost expenditures during project implementation. Implementation (Execution) Phase During the third phase, the implementation phase, the project plan is put into motion and the work of the project is performed. It is important to maintain control and communicate as needed during implementation. Progress is continuously monitored and appropriate adjustments are made and recorded as variances from the original plan. Throughout this step, project sponsors and other key stakeholders should be kept informed of the project’s status according to the agreed-on frequency and format of communication. The plan should be updated and published on a regular basis. Status reports should always emphasize the anticipated end point in terms of cost, schedule, and quality of deliverables. Each project deliverable produced should be reviewed for quality and measured against the acceptance criteria. Once all of the deliverables have been produced and the customer has accepted the final solution, the project is ready for closure. Closing.

Title of this process is The Project Life Cycle (Phases)The pr.pdf

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The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription. because it is composed with 12 subunits and due to incapable of specific promoter recognition it can\'t initiate transcription. Solution The answer is A) Pol II, with twelve subunits on its own, is capable of initiating transcription. because it is composed with 12 subunits and due to incapable of specific promoter recognition it can\'t initiate transcription..

The answer is A) Pol II, with twelve subunits on its own, is capable.pdf

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solution: A= 1 1 0 4 3 1 2 0 5 Finding the determinant of the matrix ? det A = -3 The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse matrix was calculated the minors and cofactors of the matrix ? M1,1 = 3 1 0 5 = 15 C1,1 = (-1)1+1M1,1 = 15 M1,2 = 4 1 2 5 = 18 C1,2 = (-1)1+2M1,2 = -18 M1,3 = 4 3 2 0 = -6 C1,3 = (-1)1+3M1,3 = -6 M2,1 = 1 0 0 5 = 5 C2,1 = (-1)2+1M2,1 = -5 M2,2 = 1 0 2 5 = 5 C2,2 = (-1)2+2M2,2 = 5 M2,3 = 1 1 2 0 = -2 C2,3 = (-1)2+3M2,3 = 2 M3,1 = 1 0 3 1 = 1 C3,1 = (-1)3+1M3,1 = 1 M3,2 = 1 0 4 1 = 1 C3,2 = (-1)3+2M3,2 = -1 M3,3 = 1 1 4 3 = -1 C3,3 = (-1)3+3M3,3 = -1 Matrix of cofactors: C = 15 -18 -6 -5 5 2 1 -1 -1 Transpose of cofactor matrix: CT = 15 -5 1 -18 5 -1 -6 2 -1 Finding the inverse matrix. A-1 = CT = det A -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 B) First, let\'s add the Identity Matrix to the right of our matrix 1 1 0 1 0 0 4 3 1 0 1 0 2 0 5 0 0 1 Now, let\'s do Gauss-Jordan Elimination on our new matrix... Add (-4 * row1) to row2 1 1 0 1 0 0 0 -1 1 -4 1 0 2 0 5 0 0 1 Add (-2 * row1) to row3 1 1 0 1 0 0 0 -1 1 -4 1 0 0 -2 5 -2 0 1 Divide row2 by -1 1 1 0 1 0 0 0 1 -1 4 -1 0 0 -2 5 -2 0 1 Add (2 * row2) to row3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 3 6 -2 1 Divide row3 by 3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 1 2 -2/3 1/3 Add (1 * row3) to row2 1 1 0 1 0 0 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 Add (-1 * row2) to row1 1 0 0 -5 5/3 -1/3 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 The inverse matrix can now be found in the right 3 columns of our reduced row echelon matrix Here is the inverse matrix: -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 A= 1 1 0 4 3 1 2 0 5 Solution solution: A= 1 1 0 4 3 1 2 0 5 Finding the determinant of the matrix ? det A = -3 The determinant of A is non-zero, hence the inverse matrix A-1 exists. To find the inverse matrix was calculated the minors and cofactors of the matrix ? M1,1 = 3 1 0 5 = 15 C1,1 = (-1)1+1M1,1 = 15 M1,2 = 4 1 2 5 = 18 C1,2 = (-1)1+2M1,2 = -18 M1,3 = 4 3 2 0 = -6 C1,3 = (-1)1+3M1,3 = -6 M2,1 = 1 0 0 5 = 5 C2,1 = (-1)2+1M2,1 = -5 M2,2 = 1 0 2 5 = 5 C2,2 = (-1)2+2M2,2 = 5 M2,3 = 1 1 2 0 = -2 C2,3 = (-1)2+3M2,3 = 2 M3,1 = 1 0 3 1 = 1 C3,1 = (-1)3+1M3,1 = 1 M3,2 = 1 0 4 1 = 1 C3,2 = (-1)3+2M3,2 = -1 M3,3 = 1 1 4 3 = -1 C3,3 = (-1)3+3M3,3 = -1 Matrix of cofactors: C = 15 -18 -6 -5 5 2 1 -1 -1 Transpose of cofactor matrix: CT = 15 -5 1 -18 5 -1 -6 2 -1 Finding the inverse matrix. A-1 = CT = det A -5 5/3 -1/3 6 -5/3 1/3 2 -2/3 1/3 B) First, let\'s add the Identity Matrix to the right of our matrix 1 1 0 1 0 0 4 3 1 0 1 0 2 0 5 0 0 1 Now, let\'s do Gauss-Jordan Elimination on our new matrix... Add (-4 * row1) to row2 1 1 0 1 0 0 0 -1 1 -4 1 0 2 0 5 0 0 1 Add (-2 * row1) to row3 1 1 0 1 0 0 0 -1 1 -4 1 0 0 -2 5 -2 0 1 Divide row2 by -1 1 1 0 1 0 0 0 1 -1 4 -1 0 0 -2 5 -2 0 1 Add (2 * row2) to row3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 3 6 -2 1 Divide row3 by 3 1 1 0 1 0 0 0 1 -1 4 -1 0 0 0 1 2 -2/3 1/3 Add (1 * row3) to row2 1 1 0 1 0 0 0 1 0 6 -5/3 1/3 0 0 1 2 -2/3 1/3 Add (-1 * row2) to row1 1.

solutionA= 1 1 0 4 3 1 2 0 .pdf

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Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active investigation. All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy include, 1). Diseases with known genetic defects or mutations 2). Diseases with point mutations on one or more genes 3). The possibility of delivering genes to the affected tissues 4). If the substitution of defective gene correct the disease status Solution Q1). Gene therapy is an experimental approach to treat the disease by inserting therapeutic genes or changing gene expression levels in the patient’s cells. Genes are nucleic acid polymers; these therapeutic genes used in gene therapy generally encode a functional gene that replaces mutated gene. Potential treatment for single gene genetic disorders such as haemophilia, muscular dystrophy, cystic fibrosis, etc. was demonstrated. Treatment of cancer, AIDS (acquired immune deficiency syndrome), Alzheimer’s disease and rheumatoid arthritis are also under active investigation. All the diseases cannot be cured by gene therapy. Some of hte good targets of the gene therapy include, 1). Diseases with known genetic defects or mutations 2). Diseases with point mutations on one or more genes 3). The possibility of delivering genes to the affected tissues 4). If the substitution of defective gene correct the disease status.

Q1). Gene therapy is an experimental approach to treat the disease b.pdf

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Solution : Three modes of DNA replication 1) Semi conservative: => Each daughter strand is composed of one old DNA strand and one new DNA strand. 2) Conservative: => The original double helix remains as complete unit. => The new DNA double helix is produced as a single unit. => The old DNA is completly conserved. 3) Dispersive mode: =>Separation of the individual covalent phosphodiester bonds is required for this mode of replication..

Solution Three modes of DNA replication 1) Semi conservative.pdf

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package s3; // Copy paste this Java Template and save it as \"EmergencyRoom.java\" import java.util.*; import java.io.*; // write your matric number here: // write your name here: // write list of collaborators here: // year 2016 hash code: XAbyuzR78fXeaMHBdLan (do NOT delete this line) class patientlist { private String PatientName; private int emergencyLvl; public patientlist(String PatientName, int emergencyLvl) { this.PatientName = PatientName; this.emergencyLvl = emergencyLvl; } public String getPatientName() { return PatientName; } public void setPatientName(String PatientName) { this.PatientName = PatientName; } public int getEmergencyLvl() { return emergencyLvl; } public void setEmergencyLvl(int emergencyLvl) { this.emergencyLvl = emergencyLvl; } } class EmergencyRoom { // if needed, declare a private data structure here that // is accessible to all methods in this class private List patientlist; public EmergencyRoom() { // Write necessary code during construction patientlist = new ArrayList(); } void ArriveAtHospital(String patientName, int emergencyLvl) { // You have to insert the information (patientName, emergencyLvl) // into your chosen data structure if (patientName.length() > 15 || patientName.length() < 1) { System.out.println(\"patient name is either too long or too short.\ Please enter a name between 1 to 15 characters.\"); } else if (emergencyLvl > 100 || emergencyLvl < 30) { System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid level between 30 to 100.\"); } else { int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { System.out.println(\"Patient already admitted\"); break; } } if (i == patientlist.size()) { patientlist p = new patientlist(patientName.toUpperCase(), emergencyLvl); patientlist.add(p); } } } void UpdateEmergencyLvl(String patientName, int incEmergencyLvl) { // You have to update the emergencyLvl of patientName to // emergencyLvl += incEmergencyLvl // and modify your chosen data structure (if needed) int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { //System.out.println(\"Patient already admitted\"); //break; if (incEmergencyLvl > 70 || incEmergencyLvl < 0) { System.out.println(\"Emergency level is either too high or too low.\ Please enter a valid increment between 0 to 70.\"); break; } else { patientlist.get(i).setEmergencyLvl(patientlist.get(i).getEmergencyLvl() + incEmergencyLvl); } } } } void Treat(String patientName) { // This patientName is treated by the doctor // remove him/her from your chosen data structure int i; for (i = 0; i < patientlist.size(); i++) { if (patientlist.get(i).getPatientName().compareToIgnoreCase(patientName) == 0) { patientlist.remove(i); } } } String Query() { String ans = \"The emergency room is empty\"; int i; if(patientlist.isEmpty()) return ans; else{ int max=patientlist.get(0).getEme.

package s3; Copy paste this Java Template and save it as Emer.pdf

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Our body is having two line defence system against pathogens.Pathogens are micro-organisms such like virus,bactria,parasite,fungi. First line of defence system:- It is a outside defence system which protect body from the outside ,it includes skin ,tear,mucus,cilia,stomach acid,urine flow. Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to your body unless your skin get brokken.The pathogens can get enter through your nose ,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop pathogen and break down their cell wall because they contain enzyme.those that are not killed immediately are trapped in mucus swallowed. cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine flow help to flush out pathogens from the bladder.stomach acid is present in stomach and helps to kill pathogens.this is the fine defence. Second line of defence system :- once pathogen cross your first line of defence second line of defence active for example ,through the cut pathogen enter into body and develope infection then second line defence active .Through a sequence of steps called immune response ,the immune system attacks these pathogens. In the second line of deffence system is a group of cells ,tissues,organs work together to protect the body .This is called immune system. The cells involved in the immune system called white blood cells is helps to destoys the desease causing organisms.there are different types of white blood cells but function is same that protect you from the organisms. They are neutrophiles,T-helper cells,Cytotoxic cells,marcophases,B-cells. These cells work together and kill the micro organism. Tissues and Organs :- This are involved in the immune system are the lymphatic system ,lymph nodes and lymp fluid.This are having specific function. Lymphatic system is a system of thin tube present through out the body .These tubes are called lymph vessels.they contain lymph. Lymph is a fluid that contains white blood cells. lymph helps to protect body form micro organisms. This is a second line of defence system. Solution Our body is having two line defence system against pathogens.Pathogens are micro-organisms such like virus,bactria,parasite,fungi. First line of defence system:- It is a outside defence system which protect body from the outside ,it includes skin ,tear,mucus,cilia,stomach acid,urine flow. Skin acts as a barrier between pathogens and your body. it stops all microarganisms to enter in to your body unless your skin get brokken.The pathogens can get enter through your nose ,eye,mouth, but through that if pathogen passes .Nose ,eye,mouth contains mucus,tear .they stop pathogen and break down their cell wall because they contain enzyme.those that are not killed immediately are trapped in mucus swallowed. cilia is present in our windpipe.it helps to move mucus and traped pathogens from lungs.Urine flow help to flush out.

Our body is having two line defence system against pathogens.Pathoge.pdf

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Once neurons are produced, they migrate and modify to form six layered structure of cortex. The earliest formed neurons form the preplate which is then split into marginal plate and subplate by invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and ganglionic eminences. The corticular neurons exhibit radial migration and form projection neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both types arrange in inside out sequence and the early neurons form the deep layers while the late born neuron form the outer layers of cortex. Solution Once neurons are produced, they migrate and modify to form six layered structure of cortex. The earliest formed neurons form the preplate which is then split into marginal plate and subplate by invading newly formed neurons. The neurons can have two origin: cortical ventricular zone and ganglionic eminences. The corticular neurons exhibit radial migration and form projection neurons while the ganglionic neurons exhibit tangantial migration and form interneurons. Both types arrange in inside out sequence and the early neurons form the deep layers while the late born neuron form the outer layers of cortex..

Once neurons are produced, they migrate and modify to form six layer.pdf

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PDU is called Protocol Data Unit.It consists of user data and protoc.pdf

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Answer As long as the two substituents on any one of the two carbon.pdf

1. Answer: As long as the two substituents on any one of the two carbons are not the same, it would be capable of cis/trans isomerisim. ===== Explanation: In your example of A--C=C--A, since each of the carbons have an A group and a hydrogen (not shown), it is possible for isomerism (A group is different from H). In your example of B--C=C--A, it is also capable of isomerism because the two groups on each carbon are different (B group is different from H on the first carbon, A group is also different from H on the second carbon). However, if for example, you have a two B groups both attached to the left carbon (without any hydrogens), it cannot be either cis/trans. ***Do note however, that if you have more than two different groups across the double bond overall (ex. Group A, B, and C), you should use the E/Z naming system (organic chemistry) instead of cis/trans. Solution Answer: As long as the two substituents on any one of the two carbons are not the same, it would be capable of cis/trans isomerisim. ===== Explanation: In your example of A--C=C--A, since each of the carbons have an A group and a hydrogen (not shown), it is possible for isomerism (A group is different from H). In your example of B--C=C--A, it is also capable of isomerism because the two groups on each carbon are different (B group is different from H on the first carbon, A group is also different from H on the second carbon). However, if for example, you have a two B groups both attached to the left carbon (without any

2. hydrogens), it cannot be either cis/trans. ***Do note however, that if you have more than two different groups across the double bond overall (ex. Group A, B, and C), you should use the E/Z naming system (organic chemistry) instead of cis/trans.

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